Practice C hacker rank solution

Bitwise Operators in C hackerRank step by step solution

noreply@blogger.com (Unknown) — Fri, 17 Jul 2020 21:06:00 +0530

Bitwise Operators in C hackerRank step by step solution:

Welcome back, guys!
In today's post, we will look at Bitwise Operators in C hackerRank practice problem.
It requires basic knowledge of Loops and Bitwise operators.

The following table lists the Bitwise operators supported by C. Assume variable 'A' holds 60 and variable 'B' holds 13, then −

Operator	Description	Example
&	Binary AND Operator copies a bit to the result if it exists in both operands.	(A & B) = 12, i.e., 0000 1100
\|	Binary OR Operator copies a bit if it exists in either operand.	(A \| B) = 61, i.e., 0011 1101
^	Binary XOR Operator copies the bit if it is set in one operand but not both.	(A ^ B) = 49, i.e., 0011 0001
~	Binary One's Complement Operator is unary and has the effect of 'flipping' bits.	(~A ) = ~(60), i.e,. 1100 0011
<<	Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand.	A << 2 = 240 i.e., 1111 0000
>>	Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand.	A >> 2 = 15 i.e., 0000 1111

Problem Statement for Bitwise Operators :

Objective

This challenge will let you learn about bitwise operators in C.

Inside the CPU, mathematical operations like addition, subtraction, multiplication and division are done in bit-level. To perform bit-level operations in C programming, bitwise operators are used which are explained below.

Bitwise AND operator & The output of bitwise AND is 1 if the corresponding bits of two operands is 1. If either bit of an operand is 0, the result of corresponding bit is evaluated to 0. It is denoted by &.
Bitwise OR operator | The output of bitwise OR is 1 if at least one corresponding bit of two operands is 1. It is denoted by |.
Bitwise XOR (exclusive OR) operator ^ The result of bitwise XOR operator is 1 if the corresponding bits of two operands are opposite. It is denoted by .

For example, for integers 3 and 5,

3 = 00000011 (In Binary)
5 = 00000101 (In Binary)

AND operation        OR operation        XOR operation
  00000011             00000011            00000011
& 00000101           | 00000101          ^ 00000101
  ________             ________            ________
  00000001  = 1        00000111  = 7       00000110  = 6

Task
Given set

, find:

the maximum value of which is less than a given integer , where and (where ) are two integers from set .
the maximum value of which is less than a given integer , where and (where ) are two integers from set .
the maximum value of which is less than a given integer , where and (where ) are two integers from set .

Input Format

The only line contains

space-separated integers,

and

, respectively.

Constraints

Output Format

The first line of output contains the maximum possible value of .
The second line of output contains the maximum possible value of .
The second line of output contains the maximum possible value of .

Sample Input 0

5 4

Sample Output 0

2
3
3

Explanation 0

All possible values of

and

are:

The maximum possible value of that is also is , so we print on first line.
The maximum possible value of that is also is , so we print on second line.
The maximum possible value of that is also is , so we print on third line.

Solution for Bitwise operators :

Here is a breakdown of the problem:

1)We are given numbers,n(the size of our set) and k (a constraint on the numbers we choose for our output).

2)We must have a set of integers from 1 to n that we'll refer to as S.

3)We must have 2 integers, a and b, that are elements of S. These are constrained by the rule that a<b. We implement this with nested loops.

4)We must find the bitwise AND, bitwise OR, and bitwise XOR for every possible a and b. We'll refer to these as x, y, and z respectively.

5)We must print the maximum x<k, y<k, and z<k denoted by max_and,max_or, and max_xor respectively in the output.

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

//Complete the following function.

void calculate_the_maximum(int n, int k) {

int i,j,max_and=0,max_or=0,max_xor=0;

for(i=1;i<=n;i++)

{

for(j=i+1;j<=n;j++)

{

int x=i&j,y=i|j,z=i^j;

if(x<k & x>max_and)

max_and=x;

if(y<k & y>max_or)

max_or = y;

if(z<k & z>max_xor)

max_xor = z;

}

printf("%d\n%d\n%d",max_and,max_or,max_xor);

}

int main() {

int n, k;

scanf("%d %d", &n, &k);

calculate_the_maximum(n, k);

return 0;

}

Hope it adds knowledge to you.

Analysis the code properly.

Feel Free to post your doubts in the comments down below.

See you next time.

Sum of Digits of a Five Digit Number || Hackerrank step by step solution

noreply@blogger.com (Unknown) — Fri, 26 Jun 2020 15:36:00 +0530

Sum of Digits of a Five Digit Number || Hackerrank step by step solution:

Welcome back, guys! So we will discuss Sum of Digits of a Five Digit Number Hackerrank Problem it's easy problem only knowledge of loops is required for it along with how to extract digits from a number.

Problem Statement of Sum of Digits of a Five Digit Number:

Objective

In order to get the last digit of a number, we use modulo operator \%. When the number is modulo divided by 10 we get the last digit.

To find the first digit of a number we divide the given number by 10 until the number is greater than 10. In the end, we are left with the first digit.

Task

In this challenge, you have to input a five-digit number and print the sum of digits of the number.

Input Format

The input contains a single five-digit number, n.

Constraints

10000<= n <= 99999

Output Format

Print the sum of the digits of the five-digit number.

Sample Input 0

10564

Sample Output 0

Solution for Sum of Digits of a Five Digit Number :

The modulus operator (%) comes in handy in this challenge. So how does it help? When you do a number % 10, it gives the unit's digit of the number(Why?). Then, dividing the number by 10 will eliminate the unit's digit in the number. Using a do..while loop, we can solve this challenge.

A do..while loop executes at least once because the statements in the do part of the loop are executed at least once, after which the condition in the while is checked.

What would be the condition in the while loop? Well, we want to find the sum of all the digits of the given number, so the condition in the while statement would be number != 0, which means, we need to keep executing the statements do, while the number is not equal to 0. Look at the code below to see how to solve this challenge. I will provide a solution using each loop.

Using For Loop:

#include<stdio.h>

int main()

{

int i=0,n=0,sum=0;

scanf("%d",&n);

for(i=n;i>0;i =i/10)

{ sum +=i%10;

}

printf("%d",sum);

return 0;

}

Using While Loop:

#include<stdio.h>

int main()

{

int n=0,sum=0;

scanf("%d",&n);

while(n>0)

{

sum += n%10;

n = n/10;

}

printf("%d",sum);

return 0;

}

Using DO WHILE Loop:

#include<stdio.h>

int main()

{

int n=0,sum=0;

scanf("%d",&n);

#include<stdio.h>

int main()

{

int n=0,sum=0;

scanf("%d",&n);

do{

sum += n%10;

n = n/10;

}while(n>0);

printf("%d",sum);

return 0;

}

printf("%d",sum);

return 0;

}

Analysis the code properly.

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See you Next time.

For loop in C | Hackerrank Solution

noreply@blogger.com (Unknown) — Thu, 18 Jun 2020 22:22:00 +0530

For loop in C | Hackerrank solution

Welcome Back, Guys !!
Today in this post we will discuss For loop in C Hackerrank Practice problem in C programming language.
This problem is set for students to get an introductory knowledge of using For Loop in C.

For Loop in C Problem Statement:

Objective

In this challenge, you will learn the usage of the for loop, which is a programming language statement that allows code to be repeatedly executed.

The syntax for this is

for ( <expression_1> ; <expression_2> ; <expression_3> )
<statement>
expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop.
expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
expression_3 is generally used to update the flags/variables.
A sample loop will be

for(int i = 0; i < 10; i++) {
...
}

Task

For each integer n in the interval [a,b] (given as input) :

If 1<=n<=9, then print the English representation of it in lowercase. That is "one" for, "two" for , and so on.
Else if n>9 and it is an even number, then print "even".
Else if n>9 and it is an odd number, then print "odd".
Input Format

The first line contains an integer,a .
The seond line contains an integer,b .

Constraints

1<= a<=b<=10^6

Output Format

Print the appropriate english representation,even, or odd, based on the conditions described in the 'task' section.

Sample Input

8
11

Sample Output

eight
nine
even
odd

For loop in C | Hackerrank solution

Given two numbers and, for every number starting from a till b, you have to print the English representation if the number is less than or print if it's even or odd. So, basically, you need to loop from till and print as per the given conditions. Code is as follows

Simple code using an if-else ladder:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
int a=0,b=0,n=0;
char* arr[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
scanf("%d",&a);
scanf("%d",&b);

for(n = a;n<=b;n++)
{
if(n == 1) {

printf("one\n");

}

else if(n == 2) {

printf("two\n");

}

else if(n == 3) {

printf("three\n");

}

else if(n == 4) {

printf("four\n");

}

else if(n == 5) {

printf("five\n");

}

else if(n == 6) {

printf("six\n");

}

else if(n == 7) {

printf("seven\n");

}

else if(n == 8) {

printf("eight\n");

}

else if(n == 9) {

printf("nine\n");

}

else {

if(n %2 ==0)
printf("even\n");
else
printf("odd\n");

}
}
return 0;

}

Code using arrays:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
int a=0,b=0,n=0;
char* arr[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
scanf("%d",&a);
scanf("%d",&b);

for(n = a;n<=b;n++)
{
if(n>9)
{
if(n%2 ==0)
printf("even\n");
else {
printf("odd\n");
}
}
else {
puts(arr[n]);
}
}
return 0;

}

Analysis the code properly.

Feel Free to post your doubts in the comments down below.

See you next time.

Conditional Statements in C - Hackerrank step by step solution

noreply@blogger.com (Unknown) — Wed, 3 Jun 2020 16:21:00 +0530

Conditional Statements in C - Hackerrank solution

Welcome Back, Guys !!
Today in this post we will discuss Conditional statements in C Hackerrank Practice problem in C programming language.
This problem is set for students to get better knowledge to handle conditional statement i.e if-else statement.

Conditional Statements in C - Problem statement:

Objective

if and else are two of the most frequently used conditionals in C/C++, and they enable you to execute zero or one conditional statement among many such dependent conditional statements. We use them in the following ways:

if: This executes the body of bracketed code starting with statement1 if condition evaluates to true.

if (condition) {
statement1;
...
}
if - else: This executes the body of bracketed code starting with statement1 if condition evaluates to true, or it executes the body of code starting with statement2 if condition evaluates to false. Note that only one of the bracketed code sections will ever be executed.

if (condition) {
statement1;
...
}
else {
statement2;
...
}
if - else if - else: In this structure, dependent statements are chained together and the for each statement is only checked if all prior conditions in the chain are evaluated to false. Once a evaluates to true, the bracketed code associated with that statement is executed and the program then skips to the end of the chain of statements and continues executing. If each condition in the chain evaluates to false, then the body of bracketed code in the else block at the end is executed.

if(first condition) {
...
}
else if(second condition) {
...
}
.
.
.
else if((n-1)'th condition) {
....
}
else {
...
}

Task

Given a positive integer denoting n, do the following:

If 1<=n<=9, then print the lowercase English word corresponding to the number (e.g., one for 1, two for 2, etc.).
If n>9, print Greater than 9.
Input Format

The first line contains a single integer denoting n.

Constraints
1<= n<= 10^9
Output Format

If 1<=n<=9, then print the lowercase English word corresponding to the number (e.g., one for 1, two for 2, etc.); otherwise, print Greater than 9 instead.

Sample Input

5
Sample Output

five
Sample Input #01

8
Sample Output #01

eight
Sample Input #02

44
Sample Output #02

Greater than 9

Conditional Statements in C - solution code:

Given a number, if it is less than 10, you have to print it's English representation if it is less than 10 else print "Greater than 9". This can be solved using if-else loop.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int n;
scanf("%d", &n);
if(n == 1) {
printf("one");
}
else if(n == 2) {
printf("two");
}
else if(n == 3) {
printf("three");
}
else if(n == 4) {
printf("four");
}
else if(n == 5) {
printf("five");
}
else if(n == 6) {
printf("six");
}
else if(n == 7) {
printf("seven");
}
else if(n == 8) {
printf("eight");
}
else if(n == 9) {
printf("nine");
}
else {
printf("Greater than 9");
}

return 0;

}

The code can be made short by using arrays let me show you how,

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int n;
char* a[10] = {"Greater than 9", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

scanf("%d", &n);

if(n > 9)
puts(a[0]);

else
puts(a[n]);


return 0;

}
I hope it added valuable knowledge to you.
Feel free to share your doubts in the comment section below.
See you next time.

Pointers in C - Hackerrank problem step by step solution.

noreply@blogger.com (Unknown) — Sat, 23 May 2020 16:55:00 +0530

Pointers in C - Hackerrank problem step by step solution.:

Welcome Back, Guys !!
Today in this post we will discuss Pointers in C Hackerrank Practice problem in C programming language.
This problem is set for students to get better knowledge to handle Pointers in C, and how
to use call by reference for a function call.

Pointers in C - Hackerrank problem statement:

Objective

In this challenge, you will learn to implement the basic functionalities of pointers in C. A pointer in C is a way to share a memory address among different contexts (primarily functions). They are primarily used whenever a function needs to modify the content of a variable, of which it doesn't have ownership.

In order to access the memory address of a variable,val, we need to prepend it with & sign. E.g., &val returns the memory address of val.

This memory address is assigned to a pointer and can be shared among various functions. E.g. int *p= &val will assign the memory address of val to pointer p. To access the content of the memory to which the pointer points, prepend it with a *. For example, *p will return the value reflected by val and any modification to it will be reflected at the source (val).

void increment(int *v) {
(*v)++;
}
int main() {
int a;
scanf("%d", &a);
increment(&a);
printf("%d", a);
return 0;
}
Task

You have to complete the function void update(int *a,int *b), which reads two integers as argument, and sets a with the sum of them, and b with the absolute difference of them.
a' = a+b
b' = |a-b|

Input Format

The input will contain two integers, a and b, separated by a newline.

Output Format

You have to print the updated value of a and b, on two different lines.

Note: Input/output will be automatically handled. You only have to complete the function described in the 'task' section.

Sample Input

4
5
Sample Output

9
1
Explanation

a' = 4+5 = 9
b' = |4-5|=1

Pointers in C - Hackerrank problem solution:

The explanation for the problem is given as comments in codes.

Method 1:

#include <stdio.h>
#include <stdlib.h>

void update(int *a,int *b) {
int sum=0,diff=0;
sum =(*a)+(*b);
diff=abs((*a)-(*b));
*a =sum;
*b = diff;

}

int main() {
int a, b;
int *pa = &a, *pb = &b;
// pa will have the value of address of a
// *pa will contain the value of a

scanf("%d %d", &a, &b);
update(pa, pb);
printf("%d\n%d", a, b);

return 0;
}

Method 2:

#include <stdio.h>

#include <stdlib.h>

void update(int *a,int *b) {

int k= *a;

*a =(*a)+(*b);

*b =abs((k)-(*b));

}

int main() {

int a, b;

int *pa = &a, *pb = &b;

// pa will have the value of address of a

// *pa will contain the value of a

scanf("%d %d", &a, &b);

update(pa, pb);

printf("%d\n%d", a, b);

return 0;

}

Method 3:

#include <stdio.h>

#include <stdlib.h>

void update(int *a,int *b) {

*a =(*a)+(*b);

*b =abs((*a) - 2*(*b));//note * after 2 indicates multiplication

// update a is a+b so subracting b 2 times gives the difference between them

}

int main() {

int a, b;

int *pa = &a, *pb = &b;

// pa will have the value of address of a

// *pa will contain the value of a

scanf("%d %d", &a, &b);

update(pa, pb);

printf("%d\n%d", a, b);

return 0;

}

Hope this added some valuable knowledge to you.

Post your doubts in the comment section.

See you next time.

Functions in C - Hackerrank Solution

noreply@blogger.com (Unknown) — Mon, 18 May 2020 11:33:00 +0530

Functions in C - Hackerrank Solution :

Welcome Back Guys !!
Today in this post we will discuss Functions in C HAckerrank Practice problem in C programming language.
This problem is set for students to get better knowledge to handle Functions in C.

video solution for problem

Problem Statement For Functions in C hackerrank problem:

Objective

In this challenge, you will learn simple usage of functions in C. Functions are a bunch of statements glued together. A function is provided with zero or more arguments, and it executes the statements on it. Based on the return type, it either returns nothing (void) or something.

A sample syntax for a function is

return_type function_name(arg_type_1 arg_1, arg_type_2 arg_2, ...) {
...
...
...
[if return_type is non void]
return something of type `return_type`;
}
For example, a function to read four variables and return the sum of them can be written as

int sum_of_four(int a, int b, int c, int d) {
int sum = 0;
sum += a;
sum += b;
sum += c;
sum += d;
return sum;
}

Task

You have to write a function int max_of_four(int a, int b, int c, int d) which reads four arguments and returns the greatest of them.

+= : Add and assignment operator. It adds the right operand to the left operand and assigns the result to the left operand.

a += b is equivalent to a = a + b;

Input Format

Input will contain four integers - a,b,c,d, one in each line.

Output Format

Print the greatest of the four integers.
Note: I/O will be automatically handled.

Sample Input

3
4
6
5

Sample Output

Solution Code for Functions in C Hackerrank problem:

Here we are required to write a function which take for int parameter and returns only one int value. I have used conditional operator to solve this problem . As you can see there is duplication of the code here so We can also create one more function to make our life more simple ;)

Method 1:

#include <stdio.h>

Add `int max_of_four(int a, int b, int c, int d)` here.

int max_of_four(int a,int b, int c, int d)

{

int max;

max = a>b?a:b;

max= c>max?c:max;

max = d>max?d:max;

return max;

}

int main() {

int a, b, c, d;

scanf("%d %d %d %d", &a, &b, &c, &d);

int ans = max_of_four(a, b, c, d);

printf("%d", ans);

return 0;

}

Method 2:

#include <stdio.h>

Add `int max_of_four(int a, int b, int c, int d)` here.

int max(int a, int b) {

return a > b ? a : b; // i wrote this function to avoid duplication of code.

}

int max_of_four(int a, int b, int c, int d) {

return max(a, max(b, max(c, d)));

}

int main() {

int a, b, c, d;

scanf("%d %d %d %d", &a, &b, &c, &d);

int ans = max_of_four(a, b, c, d);

printf("%d", ans);

return 0;

}

Hope This added some valueable knowledge to you.

Please read the code and try to analysis it.

See you next time :)

Playing With Characters - Hackerrank Solution

noreply@blogger.com (Unknown) — Sat, 16 May 2020 10:27:00 +0530

Playing With Characters - Hackerrank Solution :

Welcome back Guys!!

In this post we will look at the Playing With Characters Hackerrank problem.
This is an easy problem in C which makes us play with characters and how to take string
input in C programing Language.
Lets get Started...

Video Explation Of this problem...

Problem Statement for Hackerrank Problem -Playing With Characters:

Objective

This challenge will help you to learn how to take a character, a string and a sentence as input in C.

To take a single character ch as input, you can use scanf("%c", &ch ); and printf("%c", ch) writes a character specified by the argument char to stdout

char ch;

scanf("%c", &ch);

printf("%c", ch);

This piece of code prints the character ch.

You can take a string as input in C using scanf(“%s”, s). But, it accepts string only until it finds the first space.

In order to take a line as input, you can use scanf("%[^\n]%*c", s); where is defined as char s[MAX_LEN] where is the maximum size of MAX_LEN. Here, [] is the scanset character. ^\n stands for taking input until a newline isn't encountered. Then, with this %*c, it reads the newline character and here, the used * indicates that this newline character is discarded.

Note: After inputting the character and the string, inputting the sentence by the above mentioned statement won't work. This is because, at the end of each line, a new line character (\n) is present. So, the statement: scanf("%[^\n]%*c", s); will not work because the last statement will read a newline character from the previous line. This can be handled in a variety of ways and one of them being: scanf("\n"); before the last statement.

Task

You have to print the character,ch, in the first line. Then print s in next line. In the last line print the sentence,sen .

Input Format

First, take a character,ch as input.

Then take the string,s as input.

Lastly, take the sentence as input.

Output Format

Print three lines of output. The first line prints the character, .

The second line prints the string, .

The third line prints the sentence,sen .

Sample Input 0

Language

Welcome To C!!

Sample Output 0

Language

Welcome To C!!

Solution Code for Hackerrank Problem -Playing With Characters:

As mentioned in the problem statement, we need to input a character, a string without spaces and a sentence and print each of them in a new line.

To input a character,

, the statement is: scanf("%c", &ch);.
To input a string,

, the statement is: scanf("%s", s);.
To input a sentence,

, the statement is: scanf("%[^\n]%*c", sen);.

In the code below, we have declared

, as char s[20], here it represents that the string s, can hold a maximum of 20 characters.

NOTE though required size is 20 i have declared it to size 100 and used only one variable for string and sentence.

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

int main()

{

char ch,s[100] ;

/* Enter your code here. Read input from STDIN. Print output to STDOUT */

scanf("%c",&ch);

printf("%c\n",ch);

scanf("%s",s);

printf("%s\n",s);

scanf("\n");//to avoid new line character as input to next string

scanf("%[^\n]%*c", s);

// or gets(s); can be used

printf("%s",s);

return 0;

}

Hope this problem added some valueable knowledge.

Feel free to Share your thoughts in the comments below.

See you next time.

Day 1: Data Types - Hackerrank Solution

noreply@blogger.com (Unknown) — Fri, 15 May 2020 13:19:00 +0530

Day 1: Data Types - Hackerrank Solution

Welcome Back guys!!
Today in this post we will discuss Day 1: Data Types Hackerrank Problem Solution.
The objective of this post is to make code acquaint with basic data type like string ,int and doubles....

Problem Statement of Day 1: Data Types - Hackerrank Problem :

Objective

Today, we're discussing data types. Check out the Tutorial tab for learning materials and an instructional video!

Task

Complete the code in the editor below. The variables i,d and s are already declared and initialized for you. You must:

1. Declare 3 variables: one of type int, one of type double, and one of type String.
2. Read 3 lines of input from stdin (according to the sequence given in the Input Format section below) and initialize your variables.
3. Use the + operator to perform the following operations:
1.Print the sum of plus your int variable on a new line.
2.Print the sum of plus your double variable to a scale of one decimal place on a new line.
3.Concatenate with the string you read as input and print the result on a new line.
Note: If you are using a language that doesn't support using for string concatenation (e.g.: C), you can just print one variable immediately following the other on the same line. The string provided in your editor must be printed first, immediately followed by the string you read as input.

Input Format

The first line contains an integer that you must sum with .
The second line contains a double that you must sum with .
The third line contains a string that you must concatenate with .

Output Format

Print the sum of both integers on the first line, the sum of both doubles (scaled 1 to decimal place) on the second line, and then the two concatenated strings on the third line.

Sample Input

12
4.0
is the best place to learn and practice coding!

Sample Output

16
8.0
HackerRank is the best place to learn and practice coding!

Explanation

When we sum the integers 4 and 12, we get the integer 16.
When we sum the floating-point numbers 4.0 and 4.0, we get 8.0.
When we concatenate HackerRank with is the best place to learn and practice coding!, we get HackerRank is the best place to learn and practice coding!.

You will not pass this challenge if you attempt to assign the Sample Case values to your variables instead of following the instructions above and reading input from stdin.

Solution Code of Day 1: Data Types - Hackerrank Problem :

For this explanation, the variables provided in your editor are subscripted with a

, and the corresponding variables declared and initialized by you are subscripted with

The following variables are already declared and initialized in your editor:

, and

In the first test case, the following variables were read from stdin:

, and

Our first line of output is the result of

.
Our second line of output is the result of

.
Our third line of output is the result of

Code in C programing language:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
int i = 4;
double d = 4.0;

char s[] = "HackerRank ";
int i1;
double d1;
char s1[100]; // this is not scalable for input of unknown size

// Read inputs from stdin
scanf("%d", &i1);
scanf("%lf", &d1);
scanf("%*[\n] %[^\n]",s1);// or gets(s2);
//*[\n] this character is used to ignore the new line character entered after float input.

// Print outputs to stdout
printf("%d\n", i + i2);
printf("%.01lf\n", d + d2);

printf("%s%s", s, s2);
return 0;
}

Code in C++ programing language:

#include <iostream>
#include <iomanip>
#include <limits>

using namespace std;

int main() {
int i = 4;
double d = 4.0;
string s = "HackerRank ";
string str {};
int a;
double b{0.0};
cin>>a;
cin>>b;
cin.ignore();
// to ignore the new line character (enter key) after the double input
getline(cin,str);
cout<<a+i<<endl;
cout<<fixed<<setprecision(1)<<b+d<<endl;
cout<<s + str ;
return 0;

}

Code in python programing language:

i = 4
d = 4.0
s = 'HackerRank '
# Declare second integer, double, and String variables.
a =0
b =0
# Read and save an integer, double, and String to your variables.
a=int(input())
b= float(input())
string = input()
# Print the sum of both integer variables on a new line.
print(a+i)
# Print the sum of the double variables on a new line.
print(b +d)
# Concatenate and print the String variables on a new line
print(s + string)

# The 's' variable above should be printed first.

Hope this added valuable knowledge to you..
Feel Free to share your thoughts and doubts in the comment section below
Link to my youtube page : https://www.youtube.com/watch?v=aBZRBCx-LMg
See you next time.

Reverse Game - Hackerrank solution

noreply@blogger.com (Unknown) — Wed, 13 May 2020 20:16:00 +0530

Reverse Game - Hackerrank solution:
Welcome back Guys !!
Today we will be discussing the Reverse Game hackerrank problemof section Mathematics
It involves beautiful use of combinatorics....
Lets get started...

Problem Statement to hackerrank problem Reverse Game is as follows:

Akash and Akhil are playing a game. They have

balls numbered from

. Akhil asks Akash to reverse the position of the balls, i.e., to change the order from say, 0,1,2,3 to 3,2,1,0. He further asks Akash to reverse the position of the balls

times, each time starting from one position further to the right, till he reaches the last ball. So, Akash has to reverse the positions of the ball starting from

position, then from

position and so on. At the end of the game, Akhil will ask Akash the final position of any ball numbered

. Akash will win the game, if he can answer. Help Akash.

Input Format
The first line contains an integer

, i.e., the number of the test cases.
The next

lines will contain two integers

and

Output Format
Print the final index of ball

in the array.

Constraints

Sample Input

2
3 1
5 2

Sample Output

2
4

Explanation
For first test case, The rotation will be like this:
0 1 2 -> 2 1 0 -> 2 0 1 -> 2 0 1
So, Index of 1 will be 2.

Solution Code to hackerrank problem Reverse Game is as follows:

IF you analysis properly by tracing the permutations. You start with

0 1 2 3 ... (N-3) (N-2) (N-1)

The first operation flips the whole thing:

(N-1) (N-3) (N-2) ... 3 2 1 0

The second operation fixes the first, and flips the other N-1:

(N-1) 0 1 2 3 ... (N-2) (N-3)

The third operation fixes the first two and flips the other N-2:

(N-1) 0 (N-2) (N-3) ... 3 2 1.

The fourth operation fixes the first three and flips the other N-3:

(N-1) 0 (N-2) 1 2 3 ... (N-3).

Et cetera.

If you trace this out with a small example, you'll see that this process ends with ball number (N - 2 + N%2)/2, which should basically be the 'middle' of the original list, at the end. Note that the N%2 in this formula just handles the even and odd cases together: when N is even, the last number is (N - 2)/2 and when N is odd, the last number is (N - 1)/2.

Since every other step you reduce to the same problem with one less ball, this is amenable to induction if you wanted to write a formal proof.

For n = 5, the last number in the list is (5 - 2 + 1)/2 = 4/2 = 2.

0: 0 1 2 3 4

1: 4 3 2 1 0

2: 4 0 1 2 3

3: 4 0 3 2 1

4: 4 0 3 1 2

For n = 6, the last number in the list is (6 - 2 + 0)/2 = 4/2 = 2.

0: 0 1 2 3 4 5

1: 5 4 3 2 1 0

2: 5 0 1 2 3 4

3: 5 0 4 3 2 1

4: 5 0 4 1 2 3

5: 5 0 4 1 3 2

If you think about the resulting arrangement geometrically, it becomes pretty clear where the formula comes from. Think about the original sequence of numbers as an array or something.

Break the list somewhere so that the two pieces are either of equal size (N is even), or there is one more element on the right (N is odd).
Think of the things on the left as taking the odd positions, while the things on the right take the even positions.
Fold (like paper) the right side over the left.
Now merge the lists back together in the obvious way.

Concretely, here are two small cases.

N = 5:

0 1 2 3 4 breaks into 0 1 and 2 3 4.
Rewrite the left so it fits in the odd positions and rewrite the right so it fits in the even positions:

_ 0 _ 1 and 2 _ 3 _ 4

Flip the one on the right (keeping the spaces) and stack it on top of the one on the left:

4 _ 3 _ 2

_ 0 _ 1

Now merge them back together to get

4 0 3 1 2.

N = 6:

0 1 2 3 4 5 breaks into 0 1 2 and 3 4 5.
Rewrite the left so it fits in the odd positions and rewrite the right so it fits in the even positions:

_ 0 _ 1 _ 2 and 3 _ 4 _ 5.

Flip the one on the right (keeping the spaces) and stack it on top of the one on the left:

5 _ 4 _ 3

_ 0 _ 1 _ 2

Now merge them back together to get

5 0 4 1 3 2.



#include <iostream>

using namespace std;

int f(int N, int K) { return K < N / 2 ? 2 * K + 1 : 2 * (N - K - 1); }

int main() {

  int T;

  cin >> T;

  for (int i = 0; i < T; i++) {

    int N, K;

    cin >> N >> K;

    cout << f(N, K) << endl;

  }

  return 0;

}
I hope this problem has added valuable knowledge to you.

Please go through the solution properly.

Feel free to share your thoughts in the comment section below.

See you next time...

Sum and Difference of Two Numbers - Hackerrank solution:

noreply@blogger.com (Unknown) — Wed, 13 May 2020 12:17:00 +0530

Sum and Difference of Two Numbers - Hackerrank solution:

Welcome back Guys!!

In this post we will look at the Sum and Difference of Two Numbers Hackerrank problem.
This is an easy problem in C....

video solution for this problem

Problem Statement For the Sum and Difference of two numbers Hackerrank problem is as follows:

Objective

The fundamental data types in c are int, float and char. Today, we're discussing int and float data types.

The printf() function prints the given statement to the console. The syntax is printf("format string",argument_list);. In the function, if we are using an integer, character, string or float as argument, then in the format string we have to write %d (integer), %c (character), %s (string), %f (float) respectively.

The scanf() function reads the input data from the console. The syntax is scanf("format string",argument_list);. For ex: The scanf("%d",&number) statement reads integer number from the console and stores the given value in variable

To input two integers separated by a space on a single line, the command is scanf("%d %d", &n, &m), where

and

are the two integers.

Task

Your task is to take two numbers of int data type, two numbers of float data type as input and output their sum:

Declare variables: two of type int and two of type float.
Read lines of input from stdin (according to the sequence given in the 'Input Format' section below) and initialize your variables.
Use the and operator to perform the following operations:
- Print the sum and difference of two int variable on a new line.
- Print the sum and difference of two float variable rounded to one decimal place on a new line.

Input Format

The first line contains two integers.
The second line contains two floating point numbers.

Constraints

integer variables
float variables

Output Format

Print the sum and difference of both integers separated by a space on the first line, and the sum and difference of both float (scaled to

decimal place) separated by a space on the second line.

Sample Input

10  4
4.0 2.0

Sample Output

14   6
6.0 2.0

Explanation

When we sum the integers

and

, we get the integer

. When we subtract the second number

from the first number

, we get

as their difference.
When we sum the floating-point numbers

and

, we get

. When we subtract the second number

from the first number

, we get

as their difference.

Solution Code For the Sum and Difference of two numbers Hackerrank problem is as follows:

Its a basic problem whose objective is to acquaint the student with the use of basic mathematical operators.

The problem statement mentions how to input variables of a particular data type and also how to output variables.
We need to declare two variables of type int (%d) and two of type float (%f).
Input the variables using the scanf function and print the output using the printf function.
The tricky part here is to format the output of the floats such that the variable is rounded to one decimal place. This can be done like this: printf("%.1f", a). Here the .1 in the format specifier, specifies that the variable should be rounded down to 1 decimal place.

Note : Using 0.1 and .1 in format specifier makes no difference.

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

int main()

{

int a,b;

float x,y;

scanf("%d%d",&a,&b);

scanf("%f%f",&x,&y);

printf("%d %d \n",a+b,a-b);

printf("%0.1f %0.1f",x+y,x-y);

return 0;

}

I hope it added some knowledge to you.

Feel free to share your thoughts and doubts in the comment section below.

See you in the next time.

Hackerrank Solution - Hello world

noreply@blogger.com (Unknown) — Wed, 13 May 2020 11:42:00 +0530

Hackerrank Solution - Hello world:

Welcome back Guys!!

In this post we will solve Hello world Hackerrank problem

It is an indtroductory problem in C language....

Problem Statement For the Hello World Hackerrank problem:

Objective

In this challenge, we will learn some basic concepts of C that will get you started with the language. You will need to use the same syntax to read input and write output in many C challenges.

Task

This challenge requires you to print

on a single line, and then print the already provided input string to stdout.

Note: You do not need to read any input in this challenge.

Input Format

You do not need to read any input in this challenge.

Output Format

on the first line, and the string from the given input on the second line.

Sample Input 0

Welcome to C programming.

Sample Output 0

Hello, World!
Welcome to C programming.

Solution code For the Hello World Hackerrank problem:




 Problem with scanf is that as soon as it encounters space while reading a string it


stops taking input so we either use scanf("%[^\n]%*c", &s) or can use gets function;


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() 
{
 
    char s[100];
    scanf("%[^\n]%*c", &s);
    // gets(s) another alternative to read string
    
   
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    printf ("Hello, World! \n" "%s",s);   
    return 0;
}


Hope this added some knowledge and helped you.


Feel free to share your thoughts and doubts in the comment section below


Check out my youtube video as well.


See you next time.

Hackerrank Solution- Small Triangles, Large Triangles

noreply@blogger.com (Unknown) — Mon, 11 May 2020 13:31:00 +0530

Hackerrank Solution- Small Triangles, Large Triangles:

Welcome back Guys!!

In this post we will solve Small Triangles, Large Triangles Hackerrank problem.

It is a medium level problem using concept of Structure in C ...

Problem Statement for Small Triangles, Large Triangles hackerrank problem:

You are given

triangles, specifically, their sides

and

. Print them in the same style but sorted by their areas from the smallest one to the largest one. It is guaranteed that all the areas are different.

The best way to calculate a volume of the triangle with sides

and

is Heron's formula:

where

Input Format

First line of each test file contains a single integer

lines follow with

and

on each separated by single spaces.

Constraints

, and

Output Format

Print exactly

lines. On each line print

integers separated by single spaces, which are

and

of the corresponding triangle.

Sample Input 0

Sample Output 0

3 4 5
5 12 13
7 24 25

Explanation 0

The square of the first triangle is

. The square of the second triangle is

. The square of the third triangle is

. So the sorted order is the reverse one.

Solution code for Small Triangles, Large Triangles hackerrank problem:

1) Number of triangles is very small, so there is no need to use overcomplicated qsort function - you can write simple bubble or insertion sort and it will work.

2) We need to calculate volumes of triangles. But we don't need volumes themselves, only their relative comparison. Let's make some transformations:

Let's remember now that

. Now, if we substitute, we obtain

and

That means, instead of comparing volumes, we can compare

of each triangle without dealing with non-integer values at all.

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

struct triangle

{

int a;

int b;

int c;

};

typedef struct triangle triangle;

void sort_by_area(triangle *tr, int n) {

// Sort an array a of the length n

int *p=malloc(n*sizeof(int));

triangle t;

//create array of size n to store "volumes"

for(int i=0;i<n;i++)

{

float a=(tr[i].a+tr[i].b+tr[i].c)/2.0;

//use 2.0 compulsary int/int gives int, int/float gives float

p[i]=(a*(a-tr[i].a)*(a-tr[i].b)*(a-tr[i].c));

//formula without sqrt as areas are different guarenteed

//because sqrt dosent work well with float values

}

//bubble sort

for(int i=0;i<n;i++)

{

for(int j=0;j<n-i-1;j++)

{

if(p[j]>p[j+1])

{

int temp=p[j];

p[j]=p[j+1];

p[j+1]=temp;

//swapping array of areas in ascending

//and simuntaneously the structure contents

t=tr[j];

tr[j]=tr[j+1];

tr[j+1]=t;

}

int main()

{

int n;

scanf("%d", &n);

triangle *tr = malloc(n * sizeof(triangle));

for (int i = 0; i < n; i++) {

scanf("%d%d%d", &tr[i].a, &tr[i].b, &tr[i].c);

}

sort_by_area(tr, n);

for (int i = 0; i < n; i++) {

printf("%d %d %d\n", tr[i].a, tr[i].b, tr[i].c);

}

return 0;

}

Please read the code and analysis it properly.

Feel free to share your thoughts and doubts in the comment section below.

See you next time .

Variadic functions in C Hackerrank step by step solution

noreply@blogger.com (Unknown) — Mon, 11 May 2020 13:17:00 +0530

Hackerrank Solution - Variadic functions in C:

Welcome back Guys!!

In this post we will solve Variadic Function in C Hackerrank problem.

It is a medium level problem using concept of Variadic Function ...

What is variadic Function:

We typically use variadic function when we don’t know the total number of arguments that will be used for a function. Basically one single function could potentially have n number of arguments.

Problem Statement to Variadic functions in C hackerrank problem is as follows:

Variadic functions are functions which take a variable number of arguments. In C programming, a variadic function will contribute to the flexibility of the program that you are developing.

The declaration of a variadic function starts with the declaration of at least one named variable, and uses an ellipsis as the last parameter, e.g.

int printf(const char* format, ...);

In this problem, you will implement three variadic functions named

and

to calculate sums, minima, maxima of a variable number of arguments. The first argument passed to the variadic function is the count of the number of arguments, which is followed by the arguments themselves.

Input Format

The first line of the input consists of an integer .
Each test case tests the logic of your code by sending a test implementation of 3, 5 and 10 elements respectively.
You can test your code against sample/custom input.
The error log prints the parameters which are passed to the test implementation. It also prints the sum, minimum element and maximum element corresponding to your code.

Constraints

Output Format

"Correct Answer" is printed corresponding to each correct execution of a test implementation."Wrong Answer" is printed otherwise.

Sample Input 0

Sample Output 0

Correct Answer
Correct Answer
Correct Answer

Solution code for Variadic functions in C hackerrank problem is as follows:

There is no built in way for a variadic function to know how many parameters have been passed. You don't have to use them all, but if you request too many, you will get garbage. The variadic functions in this problem use a standard method of passing the number of parameters to follow as the first parameter of the function. At least one fixed parameter must precede the elipsis so two problems are solved with

variable.

So, how does one use the variable list of arguments? First declare a variable of type va_list which is called ap in the Tester's Code below. Next initialize the list with it's length (count) as va_start(ap, count). Call the generator va_arg(ap, int) telling it source and data type to retrieve the next argument in the list. Finally, when the list has been processed, it is good form to use va_end(ap).

#include <stdarg.h>

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

#define MIN_ELEMENT 1

#define MAX_ELEMENT 1000000

int sum (int count,...) {

va_list ap;

int i, sum;

va_start (ap, count); /* Initialize the argument list. */

sum = 0;

for (i = 0; i < count; i++)

sum += va_arg (ap, int); /* Get the next argument value. */

va_end (ap); /* Clean up. */

return sum;

}

int min(int count,...) {

va_list ap;

int i,m,a[count];

va_start(ap,count);

for(i=0;i<count;i++)

{

a[i]=va_arg(ap,int);

}

m=a[0];

for(i=0;i<count;i++)

{

if(a[i]<m)

{

m=a[i];

}

return m;

}

int max(int count,...) {

va_list ap;

int i,m,a[count];

va_start(ap,count);

for(i=0;i<count;i++)

{

a[i]=va_arg(ap,int);

}

m=a[0];

for(i=0;i<count;i++)

{

if(a[i]>m)

{

m=a[i];

}

return m;

}

int test_implementations_by_sending_three_elements() {

srand(time(NULL));

int elements[3];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following three elements:\n");

for (int i = 0; i < 3; i++) {

fprintf(stderr, "%d\n", elements[i]);

}

int elements_sum = sum(3, elements[0], elements[1], elements[2]);

int minimum_element = min(3, elements[0], elements[1], elements[2]);

int maximum_element = max(3, elements[0], elements[1], elements[2]);

fprintf(stderr, "Your output is:\n");

fprintf(stderr, "Elements sum is %d\n", elements_sum);

fprintf(stderr, "Minimum element is %d\n", minimum_element);

fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;

for (int i = 0; i < 3; i++) {

if (elements[i] < minimum_element) {

return 0;

}

if (elements[i] > maximum_element) {

return 0;

}

expected_elements_sum += elements[i];

}

return elements_sum == expected_elements_sum;

}

int test_implementations_by_sending_five_elements() {

srand(time(NULL));

int elements[5];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following five elements:\n");

for (int i = 0; i < 5; i++) {

fprintf(stderr, "%d\n", elements[i]);

}

int elements_sum = sum(5, elements[0], elements[1], elements[2], elements[3], elements[4]);

int minimum_element = min(5, elements[0], elements[1], elements[2], elements[3], elements[4]);

int maximum_element = max(5, elements[0], elements[1], elements[2], elements[3], elements[4]);

fprintf(stderr, "Your output is:\n");

fprintf(stderr, "Elements sum is %d\n", elements_sum);

fprintf(stderr, "Minimum element is %d\n", minimum_element);

fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;

for (int i = 0; i < 5; i++) {

if (elements[i] < minimum_element) {

return 0;

}

if (elements[i] > maximum_element) {

return 0;

}

expected_elements_sum += elements[i];

}

return elements_sum == expected_elements_sum;

}

int test_implementations_by_sending_ten_elements() {

srand(time(NULL));

int elements[10];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[5] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[6] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[7] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[8] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

elements[9] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following ten elements:\n");

for (int i = 0; i < 10; i++) {

fprintf(stderr, "%d\n", elements[i]);

}

int elements_sum = sum(10, elements[0], elements[1], elements[2], elements[3], elements[4],

elements[5], elements[6], elements[7], elements[8], elements[9]);

int minimum_element = min(10, elements[0], elements[1], elements[2], elements[3], elements[4],

elements[5], elements[6], elements[7], elements[8], elements[9]);

int maximum_element = max(10, elements[0], elements[1], elements[2], elements[3], elements[4],

elements[5], elements[6], elements[7], elements[8], elements[9]);

fprintf(stderr, "Your output is:\n");

fprintf(stderr, "Elements sum is %d\n", elements_sum);

fprintf(stderr, "Minimum element is %d\n", minimum_element);

fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;

for (int i = 0; i < 10; i++) {

if (elements[i] < minimum_element) {

return 0;

}

if (elements[i] > maximum_element) {

return 0;

}

expected_elements_sum += elements[i];

}

return elements_sum == expected_elements_sum;

}

int main ()

{

int number_of_test_cases;

scanf("%d", &number_of_test_cases);

while (number_of_test_cases--) {

if (test_implementations_by_sending_three_elements()) {

printf("Correct Answer\n");

} else {

printf("Wrong Answer\n");

}

if (test_implementations_by_sending_five_elements()) {

printf("Correct Answer\n");

} else {

printf("Wrong Answer\n");

}

if (test_implementations_by_sending_ten_elements()) {

printf("Correct Answer\n");

} else {

printf("Wrong Answer\n");

}

return 0;}

Please read the code and analysis it . It's a big one I know it will take time to grasp.

Feel free to leave your comments and doubts in the comment section below.

See you next time.

Hackerrank solution - Permutations of Strings

noreply@blogger.com (Unknown) — Mon, 11 May 2020 12:55:00 +0530

Hackerrank solution - Permutations of Strings:

Welcome back Guys!!

In this post we will solve Permutation of strings Hackerrank problem.

It is a medium level problem using concept of Character Array and String functions
from string.h library ...

The problem statement to Permutations of Strings hackerrank problem is as follows:

Strings are usually ordered in lexicographical order. That means they are ordered by comparing their leftmost different characters. For example,

because

. Also

because

. If one string is an exact prefix of the other it is lexicographically smaller, e.g.,

Given an array of strings sorted in lexicographical order, print all of its permutations in strict lexicographical order. If two permutations look the same, only print one of them. See the 'note' below for an example.

Complete the function next_permutation which generates the permutations in the described order.

For example,

. The six permutations in correct order are:

ab bc cd
ab cd bc
bc ab cd
bc cd ab
cd ab bc
cd bc ab

Note: There may be two or more of the same string as elements of

.
For example,

. Only one instance of a permutation where all elements match should be printed. In other words, if

, then print either

but not both.

A three element array having three discrete elements has six permutations as shown above. In this case, there are three matching pairs of permutations where

and

are switched. We only print the three visibly unique permutations:

ab ab bc
ab bc ab
bc ab ab

Input Format

The first line of each test file contains a single integer

, the length of the string array

Each of the next

lines contains a string

Constraints

contains only lowercase English letters.

Output Format

Print each permutation as a list of space-separated strings on a single line.

Sample Input 0

2
ab
cd

Sample Output 0

ab cd
cd ab

Sample Input 1

3
a
bc
bc

Sample Output 1

a bc bc
bc a bc
bc bc a

Explanation 1

This is similar to the note above. Only three of the six permutations are printed to avoid redundancy in output.

The solution code for Permutations of Strings hackerrank problem is as follows:

Generating the permutations in lexicographical order is a well-known problem with solution described here. The hardest part could be comparing the strings using C functions. In order to do that, we have function

That said, string

is considered to be less than

if and only if

. Notice that if

then

. So in swap function we just swap the smaller one with bigger one in terms of lexiographic order.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void swap(char **s, int i, int j){
char * tmp = s[i];
s[i] = s[j];
s[j] = tmp;
}

void reverse(char **s, int start, int end){
while(start<end){
swap(s,start++,end--);
}
}

int next_permutation(int n, char **s)
{
for(int i=n-2;i>-1;i--){
if(strcmp(s[i+1],s[i])>0){
//get min max
for(int j=n-1;j>i;j--){
if(strcmp(s[j],s[i])>0){
//do swap
swap(s,i,j);
// do reverse
reverse(s,i+1,n-1);
return 1;
}
}
}
}
return 0;
}
int main()
{
char **s;
int n;
scanf("%d", &n);
s = calloc(n, sizeof(char*));
for (int i = 0; i < n; i++)
{
s[i] = calloc(11, sizeof(char));
scanf("%s", s[i]);
}
do
{
for (int i = 0; i < n; i++)
printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
} while (next_permutation(n, s));
for (int i = 0; i < n; i++)
free(s[i]);
free(s);
return 0;

}

Please read the code and analysis it.
Feel free to share your thoughts and doubts in the comment section below.
See you next time.

Hackerrank solution - Digit Frequency

noreply@blogger.com (Unknown) — Mon, 11 May 2020 12:42:00 +0530

Hackerrank solution - Digit Frequency:

Welcome back Guys!!

In this post we will solve Digit fequency Hackerrank problem.

It is a medium level problem using concept of Character Array and ASCII value ...

Problem statement for Digit Frequency Hackerrank problem is as follows:

Given a string,

, consisting of alphabets and digits, find the frequency of each digit in the given string.

Input Format

The first line contains a string,

which is the given number.

Constraints

All the elements of num are made of english alphabets and digits.

Output Format

Print ten space-separated integers in a single line denoting the frequency of each digit from

Sample Input 0

a11472o5t6

Sample Output 0

0 2 1 0 1 1 1 1 0 0

Explanation 0

In the given string:

occurs two times.
and occur one time each.
The remaining digits and don't occur at all.

Sample Input 1

lw4n88j12n1

Sample Output 1

0 2 1 0 1 0 0 0 2 0

Sample Input 2

1v88886l256338ar0ekk

Sample Output 2

1 1 1 2 0 1 2 0 5 0

Solution Code for Digit Frequency Hackkerrank problem is as follows :

In this problem I have used only scanf ("%s",s) as theres no space in the given string. Now we know the ASCII value of 0 is 48 and that of 9 is 57 . So in this problem treating them as a charater i will search them using ASCII value and get there frequency.
You can make a frequency array and then print it but its take more time so i have avoided it.

#include<stdio.h>
#include<string.h>
int main()
{ char s[1001],i;
int f,j;
scanf("%s",s);
for(i=48;i<58;i++)
{ f=0;
for(j=0;j<strlen(s);j++)
{
if(s[j]==i)
{
f++;
}
}
printf("%d ",f);
}
return 0;

}

Please read the code and analysis it .

Feel free to share your thoughts and doubt in the comment section below.

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Hackerrank solution- Printing Tokens

noreply@blogger.com (Unknown) — Mon, 11 May 2020 12:23:00 +0530

Hackerrank solution- Printing Tokens:

Welcome back Guys!!

In this post we will solve Printing Tokens Hackerrank problem.

It is a medium level problem using concept of Character Array ...

Problem statement for Printing Tokens hackerrank Problem:

Given a sentence, s, print each word of the sentence in a new line.

Input Format

The first and only line contains a sentence, s.

Constraints

1<= len(s) <= 1000

Output Format

Print each word of the sentence in a new line.

Sample Input 0

This is C

Sample Output 0

This

is

Explanation 0

In the given string, there are three words ["This", "is", "C"]. We have to print each of these words in a new line.

Sample Input 1

Learning C is fun

Sample Output 1

Learning

is

fun

Sample Input 2

How is that

Sample Output 2

How

is

that

The soultion code For the Printing Tokens hackerrank Problem:

The logic For this problem is easy just using loop analysis the character if its a space character replace it with new line character rest keep the characters as it is...

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

char *s;
int i;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s); // or you can use gets or fgets
s = realloc(s, strlen(s) + 1);
//Write your logic to print the tokens of the sentence here.
for(i=0;i<strlen(s);i++)
{
if(s[i]==' ') // replacing the space character with new line char.
{
printf("\n");
}
else {
printf("%c",s[i]);
}
}
return 0;
}

Please read the code and analysis the code properly.

Feel free to share your thoughts and doubts in the comment section below.

See you next time.

Array reversal in c | step by step hackerrank solution

noreply@blogger.com (Unknown) — Mon, 11 May 2020 12:02:00 +0530

Hackerrank Solution - Array Reversal:

Welcome back, Guys!!

In this post, we will solve the Array Reversal Hackerrank problem.

It is a medium level problem using the concept of Array and loops...

The problem Statement For Hackerrank Array Reversal problem is as follows :

Given an array, of size , reverse it.

Example: If array, , after reversing it, the array should be, .

Input Format

The first line contains an integer, n, denoting the size of the array. The next line contains n space-separated integers denoting the elements of the array.

Constraints

1<= n <= 1000

1<= arr_i <= 1000 , where arr_i is the i^th element of the array.

Output Format

The output is handled by the code given in the editor, which would print the array.

Sample Input 0

16 13 7 2 1 12

Sample Output 0

12 1 2 7 13 16

Explanation 0

Given array, =[16, 13, 7 ,2, 1, 12] . After reversing the array, = [12 ,1 ,2 ,7 ,13 ,16 ]

Sample Input 1

1 13 15 20 12 13 2

Sample Output 1

2 13 12 20 15 13 1

Sample Input 2

15 5 16 15 17 11 5 11

Sample Output 2

11 5 11 17 15 16 5 15

The solution code for Hackerrank Array Reversal problem is as follows :

Since arrays store there values in contagious memory location i.e one after the other . So to reverse the order of array, we should start from highest index (size -1) to lowest index (i.e 0).

#include <stdio.h>

#include <stdlib.h>


int main()

{

    int num, *arr, i;

    scanf("%d", &num);

    arr = (int*) malloc(num * sizeof(int)); 

//or u can also use 'int arr[num];'or calloc func

    for(i = 0; i < num; i++) {

        scanf("%d", arr + i);

    }



    /* Write the logic to reverse the array. */

 /*

    for(i = num-1; i >= 0; i--)

        printf("%d ", *(arr + i));
it is wrong if you are doing this as you aren't reversing the array
*/
 
int s=0,e=n-1;
    while(s<e)
    {
        int t=a[s];
        a[s]=a[e];
        a[e]=t;
        s++;
        e--;
    }
   for(int i=0;i<n;i++)
   {
       printf("%d ",a[i]);
   }
    return 0;

}

Please read the code and try to anaylise it .

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1D Arrays in C step by step hackerrank solution

noreply@blogger.com (Unknown) — Mon, 11 May 2020 11:46:00 +0530

Hackerrank Solution - 1D Arrays in C:

Welcome back, Guys!!

In this post, we will solve 1D Arrays in C Hackerrank problem.

It is a medium level problem using concept of Array...

The problem statement is as follows:

An array is a container object that holds a fixed number of values of a single type. To create an array in C, we can do int arr[n];. Here, arr, is a variable array which holds up to integers. The above array is a static array that has memory allocated at compile time.
A dynamic array can be created in C, using the malloc function and the memory is allocated on the heap at runtime. To create an integer array, of size , int *arr = (int*)malloc(n * sizeof(int)), where points to the base address of the array.
In this challenge, you have to create an array of size dynamically, input the elements of the array, sum them and print the sum of the elements in a new line.

Input Format
The first line contains an integer, .
The next line contains space-separated integers.

Constraints
1<= n<= 1000
1<= a_i <= 1000

Output Format
Print in a single line the sum of the integers in the array.

Sample Input 0
6
16 13 7 2 1 12

Sample Output 0
51

Sample Input 1
7
1 13 15 20 12 13 2

Sample Output 1

Hackerrank Solution - 1D Arrays in C Solution Code:

Since this problem deals with creation of dynamic array using malloc function and suming the integers stored in the array. I advice to use calloc as it initialize the value to 0 and can be of great help. I haven't use these function in my solution but if your using it dont forget to use relloc as well.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int n,i,sum=0;
scanf("%d",&n);
int a[n];
// or you can use malloc and calloc function as well...

for(i=0;i<n;i++)
{
scanf("%d ",&a[i]);
}
for(i=0;i<n;i++)
{
sum=sum+a[i];
}
printf("%d",sum);
return 0;
}

Please read the code and try to anyalise each step properly.
Feel free to post your thoughts and doubts in the comment section below.
See you in the next post.

Printing Pattern using Loops hackerrank step by step solution

noreply@blogger.com (Unknown) — Sat, 9 May 2020 16:53:00 +0530

Printing Pattern using Loops hackerrank solution:
Welcome back, Guys!!
In this post, we will break the Printing Pattern using Loops hackerrank practice problem.
It is a medium level problem using the concept of nested loops...

Problem Statement is as follows:

In this problem, you need to print the pattern of the following form containing the numbers from to .

4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4

Input Format

The input will contain a single integer .

Constraints

1<= n<=1000

Output Format

Print the pattern mentioned in the problem statement.

Sample Input 0

2
Sample Output 0

2 2 2
2 1 2
2 2 2
Sample Input 1

5
Sample Output 1

5 5 5 5 5 5 5 5 5
5 4 4 4 4 4 4 4 5
5 4 3 3 3 3 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 2 1 2 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 5
5 5 5 5 5 5 5 5 5
Sample Input 2

7
Sample Output 2

7 7 7 7 7 7 7 7 7 7 7 7 7
7 6 6 6 6 6 6 6 6 6 6 6 7
7 6 5 5 5 5 5 5 5 5 5 6 7
7 6 5 4 4 4 4 4 4 4 5 6 7
7 6 5 4 3 3 3 3 3 4 5 6 7
7 6 5 4 3 2 2 2 3 4 5 6 7
7 6 5 4 3 2 1 2 3 4 5 6 7
7 6 5 4 3 2 2 2 3 4 5 6 7
7 6 5 4 3 3 3 3 3 4 5 6 7
7 6 5 4 4 4 4 4 4 4 5 6 7
7 6 5 5 5 5 5 5 5 5 5 6 7
7 6 6 6 6 6 6 6 6 6 6 6 7

7 7 7 7 7 7 7 7 7 7 7 7 7

Printing Pattern using Loops hackerrank solution :

There can be many alternative codes to solve this problem. I have used nested 

loops and if-else statements to solve the problem.



#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
    int i,j,n,k;
    scanf("%d",&n);
       for(i=1;i<2*n;i++)
       { k=n;
           if(i<=n)
           {
               for(j=1;j<2*n;j++)
               {
                   printf("%d ",k);
                   if(i>j)
                   {
                       k--;
                   }
                   if(i+j>=2*n)
                   {
                       k++;
                   }
               }
           }
           if(i>n)
           {for(j=1;j<2*n;j++)
               {
                   printf("%d ",k);
                   if(j>=i)
                   {
                       k++;
                   }
                   if(i+j<2*n)
                   {
                       k--;
                   }
               }
           }
           printf("\n");
       }
     return 0;
    }







Analyze the code properly.

Feel  free to share your thoughts in the comments sections , if any doubt post it in comments as well , See you next time

Dynamic Array in C hackerrank step by step solution

noreply@blogger.com (Unknown) — Fri, 8 May 2020 16:11:00 +0530

Dynamic Array in C -hackerrank solution

Welcome Guys!!
In this post, we will see Dynamic Array in C hackerrank solution.
It's one of the hard problem in hackerrank.

The Problem Statement is as follows:

Snow Howler is the librarian at the central library of the city of HuskyLand. He must handle requests which come in the following forms:

1 x y : Insert a book with pages at the end of the shelf.

2 x y : Print the number of pages in the book on the shelf.

3 x : Print the number of books on the shelf.

Snow Howler has got an assistant, Oshie, provided by the Department of Education. Although inexperienced, Oshie can handle all of the queries of types 2 and 3.

Help Snow Howler deal with all the queries of type 1.

Oshie has used two arrays:

int* total_number_of_books;
/*
* This stores the total number of books on each shelf.
*/

int** total_number_of_pages;
/*
* This stores the total number of pages in each book of each shelf.
* The rows represent the shelves and the columns represent the books.
*/
Input Format

The first line contains an integer , the number of shelves in the library.
The second line contains an integer , the number of requests.
Each of the following lines contains a request in one of the three specified formats.

Constraints

For each query of the second type, it is guaranteed that a book is present on the shelf at index.
Both the shelves and the books are numbered starting from 0.
Maximum number of books per shelf .
Output Format

Write the logic for the requests of type 1. The logic for requests of types 2 and 3 are provided.

Sample Input 0

5
5
1 0 15
1 0 20
1 2 78
2 2 0
3 0
Sample Output 0

78
2
Explanation 0

There are shelves and requests, or queries.
- 1 Place a page book at the end of shelf .
- 2 Place a page book at the end of shelf .
- 3 Place a page book at the end of shelf .
- 4 The number of pages in the book on the shelf is 78.
- 5 The number of books on the shelf is 2.

Dynamic Array in C hackerrank solution:

Dynamic arrays in C are represented by pointers with allocated memory they point on. You can use either

functions from stdlib.h but using calloc is adviced as it sets value to 0.

When a new book is added, you should increment the corresponding value from array

. However, it's not that easy for

. You should create a new array of the size

, then copy all previous values to this new array and then append

to the end of it. After all this, you should free the previously allocated memory using function

and then reassign

to this new array.

#include <stdio.h> #include <stdlib.h> /* * This stores the total number of books in each shelf. */ int* total_number_of_books; /* * This stores the total number of pages in each book of each shelf. * The rows represent the shelves and the columns represent the books. */ int** total_number_of_pages; int main() { int total_number_of_shelves; scanf("%d", &total_number_of_shelves); total_number_of_books = calloc(total_number_of_shelves, sizeof(int)); int total_number_of_queries; scanf("%d", &total_number_of_queries); total_number_of_pages = malloc(total_number_of_shelves * sizeof(int *)); for (int i = 0; i < total_number_of_shelves; i++) { total_number_of_pages[i] = calloc(1100, sizeof(int)); } while (total_number_of_queries--) { int type_of_query; scanf("%d", &type_of_query); if (type_of_query == 1) { /* * Process the query of first type here. */ int shelf, pages; scanf("%d %d", &shelf, &pages); total_number_of_books[shelf]++; int *book = total_number_of_pages[shelf]; while (*book != 0) book++; *book = pages; } else if (type_of_query == 2) { int x, y; scanf("%d %d", &x, &y); printf("%d\n", *(*(total_number_of_pages + x) + y)); } else { int x; scanf("%d", &x); printf("%d\n", *(total_number_of_books + x)); } } if (total_number_of_books) { free(total_number_of_books); } for (int i = 0; i < total_number_of_shelves; i++) { if (*(total_number_of_pages + i)) { free(*(total_number_of_pages + i)); } } if (total_number_of_pages) { free(total_number_of_pages); } return 0; }
Please read the code and anyalise it.
Feel free to share your thoughts and doubts in comment section.
See you next time.

Querying the Document hackerrank step by step solution

noreply@blogger.com (Unknown) — Fri, 8 May 2020 16:08:00 +0530

Querying the Document -hackerrank solution:

Welcome back, Guys!!
In this post, we will see Querying the Document hackerrank solution.
It's one of the hard problem in hackerrank.

The problem statement is as follows:

You will convert a raw text document into its component paragraphs, sentences and words. To test your results, queries will ask you to return a specific paragraph, sentence or word as described below.

Alicia is studying the C programming language at the University of Dunkirk and she represents the words, sentences, paragraphs, and documents using pointers:

A word is described by .

A sentence is described by . The words in the sentence are separated by one space (" "). The last word does not end with a space(" ").

A paragraph is described by . The sentences in the paragraph are separated by one period (".").

A document is described by . The paragraphs in the document are separated by one newline("\n"). The last paragraph does not end with a newline.

For example:

Learning C is fun.

Learning pointers is more fun.It is good to have pointers.

The only sentence in the first paragraph could be represented as:

char** first_sentence_in_first_paragraph = {"Learning", "C", "is", "fun"};

The first paragraph itself could be represented as:

char*** first_paragraph = {{"Learning", "C", "is", "fun"}};

The first sentence in the second paragraph could be represented as:

char** first_sentence_in_second_paragraph = {"Learning", "pointers", "is", "more", "fun"};

The second sentence in the second paragraph could be represented as:

char** second_sentence_in_second_paragraph = {"It", "is", "good", "to", "have", "pointers"};

The second paragraph could be represented as:

char*** second_paragraph = {{"Learning", "pointers", "is", "more", "fun"}, {"It", "is", "good", "to", "have", "pointers"}};

Finally, the document could be represented as:

char**** document = {{{"Learning", "C", "is", "fun"}}, {{"Learning", "pointers", "is", "more", "fun"}, {"It", "is", "good", "to", "have", "pointers"}}};

Alicia has sent a document to her friend Teodora as a string of characters, i.e. represented by not . Help her convert the document to form by completing the following functions:

to return the document represented by .

to return the paragraph.

to return the sentence in the paragraph.

to return the word in the sentence of the paragraph.

Input Format

The first line contains the integer .

Each of the next lines contains a paragraph as a single string.

The next line contains the integer , the number of queries.

Each of the next lines or groups of lines contains a query in one of the following formats:

1 The first line contains :

The next line contains an integer , the number of sentences in the paragraph.

Each of the next lines contains an integer , the number of words in the sentence.

This query corresponds to calling the function .

2 The first line contains :

The next line contains an integer , the number of words in the sentence of the paragraph.

This query corresponds to calling the function

3 The only line contains :

This query corresponds to calling the function

Constraints

The text which is passed to the has words separated by a space (" "), sentences separated by a period (".") and paragraphs separated by a newline("\n").

The last word in a sentence does not end with a space.

The last paragraph does not end with a newline.

The words contain only upper-case and lower-case English letters.

number of characters in the entire document

number of paragraphs in the entire document

Output Format

Print the paragraph, sentence or the word corresponding to the query to check the logic of your code.

Sample Input 0

Learning C is fun.

Learning pointers is more fun.It is good to have pointers.

1 2

2 1 1

3 1 1 1

Sample Output 0

Learning pointers is more fun.It is good to have pointers.

Learning C is fun

Learning

Explanation 0

The first query corresponds to returning the second paragraph with sentences of lengths and words.

The second query correspond to returning the first sentence of the first paragraph. It contains words.

The third query corresponds to returning the first word of the first sentence of the first paragraph.Querying the Document hackerrank solution
Querying the Document hackerrank solution:

The true problem here is to write function

from string. Let's write some assisting functions

and

. They all will take some substring of the original string as a parameter. You can pass the substring itself, however, in order to save time and memory, two indices are recommended instead: the index of the original string where the substring begins and the index of the original string where the substring ends.

All of these functions are written quite similar, so let's look at the pattern of writing them:

Let's find out, how many items do we have for the current container. In order to do this, let's count the number of delimeters ( '\n', '.' or ' ') in the substring provided. The number of items is equal to the number of delimeters with the notable exception for sentences where it's equal to the number of '.' exactly.
Now we can allocate memory for the container required. Both and will do the thing with the only notable difference in signature.
Let's traverse the substring one more time. This time we use a generator for the type of the items in the container (e.g. if we are creating paragraph, then we'll use for each sentence found out). Store the last index the delimeter was seen and when you meet delimeter one more time, you already know the beginning and the end of a new substring to process.

#include <stdio.h> #include <stdlib.h> #include <string.h> #include<assert.h> #define MAX_CHARACTERS 1005 #define MAX_PARAGRAPHS 5 char* kth_word_in_mth_sentence_of_nth_paragraph(char**** document, int k, int m, int n) { return document[n - 1][m - 1][k - 1]; } char** kth_sentence_in_mth_paragraph(char**** document, int k, int m) { return document[m - 1][k - 1]; } char*** kth_paragraph(char**** document, int k) { return document[k - 1]; } char**** get_document(char* text) { // doc points to an array of paragraphs char ****doc = malloc(MAX_PARAGRAPHS * sizeof(char ***)); // for each paragraph, assign an array of 1 sentence, we can reallocate later for(int i = 0; i < MAX_PARAGRAPHS; i ++) { doc[i] = malloc(1 * sizeof(char **)); } // for each sentence assign an array of 1 word for(int i = 0; i < MAX_PARAGRAPHS; i ++) { for(int j = 0; j < 1; j ++) { doc[i][j] = malloc(1 * sizeof(char*)); } } // for each word assign an array of 1 character for(int i = 0; i < MAX_PARAGRAPHS; i ++) { for(int j = 0; j < 1; j ++) { for(int k = 0; k < 1; k ++) { doc[i][j][k] = malloc(1 * sizeof(char)); } } } for(int n = 0, i = 0, j = 0, k = 0, l = 0; n < strlen(text); n ++) { if(text[n] != ' ' && text[n] != '\n' && text[n] != '.') { doc[i][j][k][l] = text[n]; l++; doc[i][j][k] = realloc(doc[i][j][k], (l + 1) * sizeof(char)); } else if(text[n] == ' ') { doc[i][j][k][l] = '\0'; l = 0; k++; doc[i][j] = realloc(doc[i][j], (k + 1) * sizeof(char*)); doc[i][j][k] = malloc(sizeof(char)); continue; } else if(text[n] == '.') { doc[i][j][k][l] = '\0'; k = l = 0; j++; doc[i] = realloc(doc[i], (j+1) * sizeof(char**)); doc[i][j] = malloc(sizeof(char*)); doc[i][j][k] = malloc(sizeof(char)); continue; } else if(text[n] == '\n') { j = k = l = 0; i++; continue; } } return doc; } char* get_input_text() { int paragraph_count; scanf("%d", &paragraph_count); char p[MAX_PARAGRAPHS][MAX_CHARACTERS], doc[MAX_CHARACTERS]; memset(doc, 0, sizeof(doc)); getchar(); for (int i = 0; i < paragraph_count; i++) { scanf("%[^\n]%*c", p[i]); strcat(doc, p[i]); if (i != paragraph_count - 1) strcat(doc, "\n"); } char* returnDoc = (char*)malloc((strlen (doc)+1) * (sizeof(char))); strcpy(returnDoc, doc); return returnDoc; } void print_word(char* word) { printf("%s", word); } void print_sentence(char** sentence) { int word_count; scanf("%d", &word_count); for(int i = 0; i < word_count; i++){ printf("%s", sentence[i]); if( i != word_count - 1) printf(" "); } } void print_paragraph(char*** paragraph) { int sentence_count; scanf("%d", &sentence_count); for (int i = 0; i < sentence_count; i++) { print_sentence(*(paragraph + i)); printf("."); } } int main() { char* text = get_input_text(); char**** document = get_document(text); int q; scanf("%d", &q); while (q--) { int type; scanf("%d", &type); if (type == 3){ int k, m, n; scanf("%d %d %d", &k, &m, &n); char* word = kth_word_in_mth_sentence_of_nth_paragraph(document, k, m, n); print_word(word); } else if (type == 2){ int k, m; scanf("%d %d", &k, &m); char** sentence = kth_sentence_in_mth_paragraph(document, k, m); print_sentence(sentence); } else{ int k; scanf("%d", &k); char*** paragraph = kth_paragraph(document, k); print_paragraph(paragraph); } printf("\n"); } }

Feel free to share your thoughts and doubts in comment section.

See you next time.

Sorting Array of Strings hackerrank C step by step solution

noreply@blogger.com (Unknown) — Fri, 8 May 2020 15:56:00 +0530

Sorting Array of Strings hackerrank solution:

Welcome back, Guys!!
In this post, we will see Sorting Array of Strings hackerrank step by step solution.
It's one of the hard problem of hackerrank.

The problem statement is as follows:

To sort a given array of strings into lexicographically increasing order or into an order in which the string with the lowest length appears first, a sorting function with a flag indicating the type of comparison strategy can be written. The disadvantage with doing so is having to rewrite the function for every new comparison strategy.

A better implementation would be to write a sorting function that accepts a pointer to the function that compares each pair of strings. Doing this will mean only passing a pointer to the sorting function with every new comparison strategy.

Given an array of strings, you need to implement a function which sorts the strings according to a comparison function, i.e, you need to implement the function :

void string_sort(const char **arr,const int cnt, int (*cmp_func)(const char* a, const char* b)){

}

The arguments passed to this function are:

an array of strings :

length of string array:

pointer to the string comparison function:

You also need to implement the following four string comparison functions:

to sort the strings in lexicographically non-decreasing order.

to sort the strings in lexicographically non-increasing order.

to sort the strings in non-decreasing order of the number of distinct characters present in them. If two strings have the same number of distinct characters present in them, then the lexicographically smaller string should appear first.

to sort the strings in non-decreasing order of their lengths. If two strings have the same length, then the lexicographically smaller string should appear first.

Input Format

You just need to complete the function string\_sort and implement the four string comparison functions.

Constraints

No. of Strings

Total Length of all the strings

You have to write your own sorting function and you cannot use the inbuilt function

The strings consists of lower-case English Alphabets only.

Output Format

The locked code-stub will check the logic of your code. The output consists of the strings sorted according to the four comparsion functions in the order mentioned in the problem statement.

Sample Input 0

wkue

qoi

sbv

fekls

Sample Output 0

fekls

qoi

sbv

wkue

sbv

qoi

fekls

qoi

sbv

wkue

fekls

qoi

sbv

wkue

fekls

Sorting Array of Strings hacker rank solution :

You can use any sorting algorithm to sort strings in

. Bubble sort is fast enough for the constraints in this problem. Let's look how should the bodies of the comparing functions look like.

For the lexicographical sort you should use

from the library

. It does exactly what the problem wants. You should not forget to use different signs in the functions:

means

is lexicographically bigger than

vice versa.

For the

you should calculate number of distinct characters in the strings. In order to do this, declare an array

of length

with all items equal to

. Go through all characters of your string and when you meet the letter with the corresponding item in

equal to

set it to

and add

to the number of distinct letters.

For the sorting by length you should use

from

. It returns exactly the length of the string you wanted

#include <stdio.h> #include <stdlib.h> #include <string.h> int lexicographic_sort(const char* a, const char* b){ return strcmp(a, b); } int lexicographic_sort_reverse(const char* a, const char* b){ return strcmp(b, a); } int characters_count(const char *s){ int n = 0; int count[128] = {0}; if (NULL == s){ return -1; } while(*s != '\0'){ if (!count[*s]){ count[*s]++; n++; } s++; } return n; } int sort_by_number_of_distinct_characters(const char* a, const char* b){ int con = characters_count(a) - characters_count(b); return (con ? con : lexicographic_sort(a, b)); } int sort_by_length(const char* a, const char* b) { int len = strlen(a) - strlen(b); return (len ? len : lexicographic_sort(a, b)); } void string_sort(char** arr,const int len,int (*cmp_func)(const char* a, const char* b)){ int mid = len / 2; int con = 0; while(!con){ con = 1; for(int i = 0; i < len - 1; i++){ if(cmp_func(arr[i], arr[i + 1]) > 0) { char *temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; con = 0; } } } } int main() { int n; scanf("%d", &n); char** arr; arr = (char**)malloc(n * sizeof(char*)); for(int i = 0; i < n; i++){ *(arr + i) = malloc(1024 * sizeof(char)); scanf("%s", *(arr + i)); *(arr + i) = realloc(*(arr + i), strlen(*(arr + i)) + 1); } string_sort(arr, n, lexicographic_sort); for(int i = 0; i < n; i++) printf("%s\n", arr[i]); printf("\n"); string_sort(arr, n, lexicographic_sort_reverse); for(int i = 0; i < n; i++) printf("%s\n", arr[i]); printf("\n"); string_sort(arr, n, sort_by_length); for(int i = 0; i < n; i++) printf("%s\n", arr[i]); printf("\n"); string_sort(arr, n, sort_by_number_of_distinct_characters); for(int i = 0; i < n; i++) printf("%s\n", arr[i]); printf("\n"); }

Feel free to share your thoughts and doubt in the comment section below.

See you next time...

Structuring the Document Hacker rank C step by step solution

noreply@blogger.com (Unknown) — Fri, 8 May 2020 15:50:00 +0530

Structuring the Document Hackerrank C solution:

Welcome Guys!!
In this post we will see Structuring the Document Hacker rank C solution.
Its one of the hard problem for subtopic Struct and Enums.

The problem Statement of

Structuring the Document Hackerrank C

is as follows:

A document is represented as a collection paragraphs, a paragraph is represented as a collection of sentences, a sentence is represented as a collection of words and a word is represented as a collection of lower-case ([a-z]) and upper-case ([A-Z]) English characters. You will convert a raw text document into its component paragraphs, sentences and words. To test your results, queries will ask you to return a specific paragraph, sentence or word as described below.

Alicia is studying the C programming language at the University of Dunkirk and she represents the words, sentences, paragraphs, and documents using pointers:

A word is described by:

struct word {

char* data;

};

A sentence is described by:

struct sentence {

struct word* data;

int word_count;//the number of words in a sentence

};

The words in the sentence are separated by one space (" "). The last word does not end with a space.

A paragraph is described by:

struct paragraph {

struct sentence* data ;

int sentence_count;//the number of sentences in a paragraph

};

The sentences in the paragraph are separated by one period (".").

A document is described by:

struct document {

struct paragraph* data;

int paragraph_count;//the number of paragraphs in a document

};

The paragraphs in the document are separated by one newline("\n"). The last paragraph does not end with a newline.

For example:

Learning C is fun.

Learning pointers is more fun.It is good to have pointers.

The only sentence in the first paragraph could be represented as:

struct sentence first_sentence_in_first_paragraph;

first_sentence_in_first_paragraph.data = {"Learning", "C", "is", "fun"};

The first paragraph itself could be represented as:

struct paragraph first_paragraph;

first_paragraph.data = {{"Learning", "C", "is", "fun"}};

The first sentence in the second paragraph could be represented as:

struct sentence first_sentence_in_second_paragraph;

first_sentence_in_second_paragraph.data = {"Learning", "pointers", "is", "more", "fun"};

The second sentence in the second paragraph could be represented as:

struct sentence second_sentence_in_second_paragraph;

second_sentence_in_second_paragraph.data = {"It", "is", "good", "to", "have", "pointers"};

The second paragraph could be represented as:

struct paragraph second_paragraph;

second_paragraph.data = {{"Learning", "pointers", "is", "more", "fun"}, {"It", "is", "good", "to", "have", "pointers"}};

Finally, the document could be represented as:

struct document Doc;

Doc.data = {{{"Learning", "C", "is", "fun"}}, {{"Learning", "pointers", "is", "more", "fun"}, {"It", "is", "good", "to", "have", "pointers"}}};

Alicia has sent a document to her friend Teodora as a string of characters, i.e. represented by not . Help her convert the document to form by completing the following functions:

to intialise the document. You have to intialise the global variable of type .

to return the paragraph in the document.

to return the sentence in the paragraph.

to return the word in the sentence of the paragraph.

Input Format

The first line contains the integer .

Each of the next lines contains a paragraph as a single string.

The next line contains the integer , the number of queries.

Each of the next lines contains a query in one of the following formats:

: This corresponds to calling the function .

Constraints

The text which is passed to has words separated by a spaces(" "), sentences separated by a period(".") and paragraphs separated by a newline("\n").

The last word in a sentence does not end with a space.

The last paragraph does not end with a newline.

The words contain only upper-case and lower-case English letters.

number of characters in the entire document .

number of paragraphs in the entire document .

Output Format

Print the paragraph, sentence or the word corresponding to the query to check the logic of your code.

Sample Input 0

Learning C is fun.

Learning pointers is more fun.It is good to have pointers.

1 2

2 1 1

3 1 1 1

Sample Output 0

Learning pointers is more fun.It is good to have pointers.

Learning C is fun

Learning

Explanation 0

The first query returns the second paragraph.

The second query returns the first sentence of the first paragraph.

The third query returns the first word of the first sentence of the first paragraph.

Structuring the Document Hacker rank C solution:

The true problem here is to write function from string. Let's write some assisting functions , and . They all will take some substring of the original string as a parameter. You can pass the substring itself, however, in order to save time and memory, two indices are recommended instead: the index of the original string where the substring begins and the index of the original string where the substring ends.

All of these functions are written quite similar, so let's look at the pattern of writing them:

Let's find out, how many items do we have for the current container. In order to do this, let's count the number of delimeters ( '\n', '.' or ' ') in the substring provided. The number of items is equal to the number of delimeters with the notable exception for sentences where it's equal to the number of '.' exactly.
Now we can allocate memory for the container required. Both and will do the thing with the only notable difference in signature.
Let's traverse the substring one more time. This time we use a generator for the type of the items in the container (e.g. if we are creating paragraph, then we'll use for each sentence found out). Store the last index the delimeter was seen and when you meet delimeter one more time, you already know the beginning and the end of a new substring to process.

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #define MAX_CHARACTERS 1005 #define MAX_PARAGRAPHS 5 struct word { char* data; }; struct sentence { struct word* data; int word_count;//denotes number of words in a sentence }; struct paragraph { struct sentence* data ; int sentence_count;//denotes number of sentences in a paragraph }; struct document { struct paragraph* data; int paragraph_count;//denotes number of paragraphs in a document }; struct document get_document(char* text) { struct document doc; struct paragraph *cur_paragraph = NULL; struct sentence *cur_sentence = NULL; char *new_word = NULL; doc.data = NULL; doc.paragraph_count = 0; for (char *s = text; *s; ++s) { if (*s == ' ' || *s == '.') { // new paragraph if (cur_paragraph == NULL) { doc.paragraph_count++; doc.data = (struct paragraph *) realloc(doc.data, sizeof(struct paragraph) * doc.paragraph_count); cur_paragraph = doc.data + doc.paragraph_count - 1; cur_paragraph->data = NULL; cur_paragraph->sentence_count = 0; cur_sentence = NULL; // we start a sentence again } // new sentence if (cur_sentence == NULL) { cur_paragraph->sentence_count++; cur_paragraph->data = (struct sentence *) realloc(cur_paragraph->data, sizeof(struct sentence) * cur_paragraph->sentence_count); cur_sentence = cur_paragraph->data + cur_paragraph->sentence_count - 1; cur_sentence->data = NULL; cur_sentence->word_count = 0; } // new word cur_sentence->word_count++; cur_sentence->data = (struct word *) realloc(cur_sentence->data, sizeof(struct word) * cur_sentence->word_count); cur_sentence->data[cur_sentence->word_count - 1].data = new_word; new_word = NULL; if (*s == '.') cur_sentence = NULL; // we will start a sentence again *s = 0; } else if (*s == '\n') { cur_sentence = NULL; cur_paragraph = NULL; } else { if (new_word == NULL) { new_word = s; } } } return doc; } struct word kth_word_in_mth_sentence_of_nth_paragraph(struct document Doc, int k, int m, int n) { return Doc.data[n - 1].data[m - 1].data[k - 1]; } struct sentence kth_sentence_in_mth_paragraph(struct document Doc, int k, int m) { return Doc.data[m - 1].data[k - 1]; } struct paragraph kth_paragraph(struct document Doc, int k) { return Doc.data[k - 1]; } void print_word(struct word w) { printf("%s", w.data); } void print_sentence(struct sentence sen) { for(int i = 0; i < sen.word_count; i++) { print_word(sen.data[i]); if (i != sen.word_count - 1) { printf(" "); } } } void print_paragraph(struct paragraph para) { for(int i = 0; i < para.sentence_count; i++){ print_sentence(para.data[i]); printf("."); } } void print_document(struct document doc) { for(int i = 0; i < doc.paragraph_count; i++) { print_paragraph(doc.data[i]); if (i != doc.paragraph_count - 1) printf("\n"); } } char* get_input_text() { int paragraph_count; scanf("%d", &paragraph_count); char p[MAX_PARAGRAPHS][MAX_CHARACTERS], doc[MAX_CHARACTERS]; memset(doc, 0, sizeof(doc)); getchar(); for (int i = 0; i < paragraph_count; i++) { scanf("%[^\n]%*c", p[i]); strcat(doc, p[i]); if (i != paragraph_count - 1) strcat(doc, "\n"); } char* returnDoc = (char*)malloc((strlen (doc)+1) * (sizeof(char))); strcpy(returnDoc, doc); return returnDoc; } int main() { char* text = get_input_text(); struct document Doc = get_document(text); int q; scanf("%d", &q); while (q--) { int type; scanf("%d", &type); if (type == 3){ int k, m, n; scanf("%d %d %d", &k, &m, &n); struct word w = kth_word_in_mth_sentence_of_nth_paragraph(Doc, k, m, n); print_word(w); } else if (type == 2) { int k, m; scanf("%d %d", &k, &m); struct sentence sen= kth_sentence_in_mth_paragraph(Doc, k, m); print_sentence(sen); } else{ int k; scanf("%d", &k); struct paragraph para = kth_paragraph(Doc, k); print_paragraph(para); } printf("\n"); } }

Analyze the code properly.

Feel free to share your thoughts and doubt in the comment section below.

Post transition - Hackerrank Solution in C language

noreply@blogger.com (Unknown) — Wed, 6 May 2020 16:16:00 +0530

Post transition - Hackerrank Solution in C language

Welcome Guys!!

In this post we will see the Post Transition HackerRank solution in C.

It's one of the hard problems for subtopic Struct and Enums.

The Link to the Problem statement of Post Transition is as follows :

https://www.hackerrank.com/challenges/post-transition/problem

Post Transition - Hacker Rank Solution:

For print_all_packages you should simply traverse every town, office and package in the correct order.

For send_all_acceptable_packages first you should determine the number of packages in two offices after transaction. After you determined it, create two dynamic arrays for the packages for sender office and for recipient one. Fill it with appropriate packages and assign to the field packages of responsive offices. Don't forget to free the old ones!

For town_with_most_packages you should create a helper function number_of_packages, that returns the number of packages in the town. After doing this, simply go through all towns and compare the number of packages in each.

#include <stdio.h>

#include <stdlib.h>

#define MAX_STRING_LENGTH 6

struct package

{

   char* id;

   int weight;

};

typedef struct package package;

struct post_office

{

   int min_weight;

   int max_weight;

   package* packages;

   int packages_count;

};

typedef struct post_office post_office;

struct town

{

   char* name;

   post_office* offices;

   int offices_count;

};

typedef struct town town;

void print_all_packages(town t)

{

    printf("%s:\n", t.name);

    for (int i = 0; i < t.offices_count; i++)

    {

        printf("\t%d:\n", i);

        for (int j = 0; j < t.offices[i].packages_count; j++)

            printf("\t\t%s\n", t.offices[i].packages[j].id);

    }

}

void send_all_acceptable_packages(town* source, int source_office_index, town* target, int target_office_index)

{

    int n = 0;

    for (int i = 0; i < source->offices[source_office_index].packages_count; i++)

        if (source->offices[source_office_index].packages[i].weight >= target->offices[target_office_index].min_weight &&

            source->offices[source_office_index].packages[i].weight <= target->offices[target_office_index].max_weight)

            ++n;

    package* newPackages = malloc(sizeof(package)(n + target->offices[target_office_index].packages_count));

    package oldPackages = malloc(sizeof(package)(source->offices[source_office_index].packages_count - n));

    for (int i = 0; i < target->offices[target_office_index].packages_count; i++)

        newPackages[i] = target->offices[target_office_index].packages[i];

    n = target->offices[target_office_index].packages_count;

    int m = 0;

    for (int i = 0; i < source->offices[source_office_index].packages_count; i++)

        if (source->offices[source_office_index].packages[i].weight >= target->offices[target_office_index].min_weight &&

            source->offices[source_office_index].packages[i].weight <= target->offices[target_office_index].max_weight)

        {

            newPackages[n] = source->offices[source_office_index].packages[i];

            ++n;

        }

        else

        {

            oldPackages[m] = source->offices[source_office_index].packages[i];

            ++m;

        }

    target->offices[target_office_index].packages_count = n;

    free(target->offices[target_office_index].packages);

    target->offices[target_office_index].packages = newPackages;

    source->offices[source_office_index].packages_count = m;

    free(source->offices[source_office_index].packages);

    source->offices[source_office_index].packages = oldPackages;

}

int number_of_packages(town t)

{

    int ans = 0;

    for (int i = 0; i < t.offices_count; i++)

        ans += t.offices[i].packages_count;

    return ans;

}

town town_with_most_packages(town towns, int towns_count)

{

    int ans;

    int max_packages = -1;

    for (int i = 0; i < towns_count; i++)

        if (number_of_packages(towns[i]) > max_packages)

        {

            max_packages = number_of_packages(towns[i]);

            ans = i;

        }

    return towns[ans];

}

town* find_town(town* towns, int towns_count, char* name)

{

    for (int i = 0; i < towns_count; i++)

        if (!strcmp(towns[i].name, name))

            return &(towns[i]);

    return &towns[0];

}

int main()

{

   int towns_count;

   scanf("%d", &towns_count);

   town* towns = malloc(sizeof(town)towns_count);

   for (int i = 0; i < towns_count; i++) {

      towns[i].name = malloc(sizeof(char) MAX_STRING_LENGTH);

      scanf("%s", towns[i].name);

      scanf("%d", &towns[i].offices_count);

      towns[i].offices = malloc(sizeof(post_office)towns[i].offices_count);

      for (int j = 0; j < towns[i].offices_count; j++) {

         scanf("%d%d%d", &towns[i].offices[j].packages_count, &towns[i].offices[j].min_weight, &towns[i].offices[j].max_weight);

         towns[i].offices[j].packages = malloc(sizeof(package)towns[i].offices[j].packages_count);

         for (int k = 0; k < towns[i].offices[j].packages_count; k++) {

            towns[i].offices[j].packages[k].id = malloc(sizeof(char) * MAX_STRING_LENGTH);

            scanf("%s", towns[i].offices[j].packages[k].id);

            scanf("%d", &towns[i].offices[j].packages[k].weight);

         }

      }

   }

   int queries;

   scanf("%d", &queries);

   char town_name[MAX_STRING_LENGTH];

   while (queries--) {

      int type;

      scanf("%d", &type);

      switch (type) {

      case 1:

         scanf("%s", town_name);

         town* t = find_town(towns, towns_count, town_name);

         print_all_packages(t);

         break;

      case 2:

         scanf("%s", town_name);

         town source = find_town(towns, towns_count, town_name);

         int source_index;

         scanf("%d", &source_index);

         scanf("%s", town_name);

         town* target = find_town(towns, towns_count, town_name);

         int target_index;

         scanf("%d", &target_index);

         send_all_acceptable_packages(source, source_index, target, target_index);

         break;

      case 3:

         printf("Town with the most number of packages is %s\n", town_with_most_packages(towns, towns_count).name);

         break;

      }

   }

   return 0;

}

Please read and go through the code and try to analysis the code .

IF you have a doubt feel free to share it in the comment section.

See you in the next post ;)