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Question 3: Wire {HDLBits-Verilog-Solutions}

noreply@blogger.com (TechinGrow) — Tue, 25 Apr 2023 02:54:00 +0000

Welcome to HDLBits-Verilog-Solutions!

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Question 3: Create a module with one input and one output that behaves like a wire.

Unlike physical wires, wires (and other signals) in Verilog are directional. This means information flows in only one direction, from (usually one) source to the sinks (The source is also often called a driver that drives a value onto a wire). In a Verilog "continuous assignment" (assign left_side = right_side;), the value of the signal on the right side is driven onto the wire on the left side. The assignment is "continuous" because the assignment continues all the time even if the right side's value changes. A continuous assignment is not a one-time event.

The ports on a module also have a direction (usually input or output). An input port is driven by something from outside the module, while an output port drives something outside. When viewed from inside the module, an input port is a driver or source, while an output port is a sink.

The diagram below illustrates how each part of the circuit corresponds to each bit of Verilog code. The module and port declarations create the black portions of the circuit. Your task is to create a wire (in green) by adding an assign statement to connect in to out. The parts outside the box are not your concern, but you should know that your circuit is tested by connecting signals from our test harness to the ports on your top_module.

Fig 3.1

Solution:

module top_module( input in, output out );

assign out=in;

endmodule

---------------------------------------------------------------------------

Thank you!!!!

Question 2: Output_Zero {HDLBits-Verilog-Solutions}

noreply@blogger.com (TechinGrow) — Tue, 25 Apr 2023 02:48:00 +0000

Welcome to HDLBits-Verilog-Solutions!

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Question 2: Build a circuit with no inputs and one output that outputs a constant 0.

Solution:

module top_module (output zero); // Module body starts after semicolon

assign zero=0;

endmodule

-----------------------------------------------------------------------------------------------------------------

Thank you!!!!

Question 1: Getting_Started {HDLBits-Verilog-Solutions}

noreply@blogger.com (TechinGrow) — Tue, 25 Apr 2023 02:44:00 +0000

Welcome to HDLBits-Verilog-Solutions!

Getting started in Digital Logic Design can be overwhelming at first because you need to learn new concepts, a new Hardware Description Language (e.g., Verilog), several new software packages, and often an FPGA board, all at the same time. HDLBits provides a way to practice designing and debugging simple circuits with a single click of "Simulate".

Designing a circuit requires several steps: Writing HDL (Verilog) code, compiling the code to produce a circuit, then simulating the circuit and fixing bugs.

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Question 1: Build a circuit with no inputs and one output. That output should always drive 1 (or logic high).

Solution:

module top_module( output one );

// Insert your code here

assign one = 1;

endmodule

-----------------------------------------------------------------------------------------------------------------

Thank you!!!!

Analysis of n-MOSFET using Cadence Virtuoso

noreply@blogger.com (TechinGrow) — Sun, 30 Jan 2022 06:32:00 +0000

In this blog we will see how to perform Analysis of n-MOSFET on Cadence Virtuoso,

Steps to perform a Analysis of n-MOSFET are as follows:

1. Go to Instance --> browse --> gpdk180 --> nmos --> symbol --> close --> hide

2. Again Instance --> browse --> analogLib --> vdc -->symbol --> close --> hide

3. Do wiring in circuit and add "gnd"

4. In the properties of both voltage source V1 and V2, in DC voltage name them as V1 as "vgs" and V2 as "vds"

5. Click on "check and save", if shows error then correct it.

6. Launch --> ADEL

7. Right click --> copy from cellview

8. Give value as 0.8 to both vgs and vds respectively

9. Analyses --> DC --> "save dc operating point" --> design variable --> select "vgs" for transfer characteristics --> range 0 to 1.4

10. Outputs --> "To be plotted" --> select the upper terminal of nmos for "Id"

11. RUN

Now, for Output characteristics of n-MOSFET

12. Go to Analyses --> choose --> change variable "vgs" to "vds"

13. RUN

I hope this will help you, if want to know more about cadence then you can visit,

Playlist: https://youtube.com/playlist?list=PLrwRwVZrfkcratatpzgtALFDhgcq9hCs8

Thanks...

Verification of Characteristics of Diode using Cadence Virtuoso

noreply@blogger.com (TechinGrow) — Sat, 29 Jan 2022 05:40:00 +0000

In this blog we will see how to perform Characteristics of Diode on Cadence Virtuoso,

1. Forward Bias

2. Reverse Bias

Steps:

1. Create instance by selecting “create instance” command or by pressing “I” on keyboard.

2. For diode, lib -> gpdk180, cell -> ndio, select “symbol” -> close -> hide

3. For resistance, lib -> analogLib, cell -> res, select “symbol” -> close -> hide

4. For ground, lib -> analogLib, cell -> gnd, select “symbol” -> close -> hide

5. For dc voltage, lib -> analogLib, cell -> vdc, select “symbol” -> close -> hide

6. Change properties of component by pressing key “q” after selecting that component.

7. Dc voltage = Vd, i.e. Vdc=Vd

8. check and save or “ctrl+S”

9. launch -> ADE L {Analog Design Environment Launch}

10. setup -> modal libraries -> select correct path of your schematic if not selected.

11. Design variables -> Vd -> 1

12. Analyses -> choose

Here set values according to condition that whether it is forward bias or reverse bias. Above example is for reverse bias.

For forward bias start->0 and stop->+1.8

Step size-> 0.1 or no. of steps as 20

13. Variables -> copy from cellview

14. Output -> plotted from cellview ->select from schematic -> select a terminal point of diode where you want to find a V-I characteristics.

15. Click on Netlist and Run.

Thanks....

Use of Tools in Cadence Virtuoso

noreply@blogger.com (TechinGrow) — Sat, 29 Jan 2022 05:23:00 +0000

Cadence Virtuoso helps in making integrated circuit design.... Cadence Virtuoso provides various features. It has very useful tools.

To know the use of tools and commands in cadence virtuoso watch the embedded video given below:

YouTube Channel Link: https://www.youtube.com/channel/UCX4XuMGYu8ksAO4Nnia1umQ

Thanks...

How to create Library and Cellview in Cadence Virtuoso ?....

noreply@blogger.com (TechinGrow) — Thu, 27 Jan 2022 18:14:00 +0000

In this tutorial, we will see how to create Library and Cellview in Cadence Virtuoso. I will mention only the steps here, if you want to see this process in detail then you can prefer the following link,

Link: https://youtu.be/lAlEWwCzrvU

*****Steps for creating Library and Cellview in Cadence Virtuoso:*****

1. Open Cadence Virtuoso.

2. Click on Virtuoso Dialog Box.

3. Click on File.

4. Click on New à Library à give name to library.

5. While making library make sure to select “Attach to an existing Library” à click “OK” à select “gpdk180” as technology.

6. “gpdk180” is simply a library in which certain rules a defined according to requirements of semiconductors for certification of schematic design of circuit.

7. Click “OK”.

8. Now, again click on Virtuoso dialog box à File à “Cellview”.

9. Give name to “Cellview” under the “Library” which you have made.

10. Click “OK”.

11. Now, schematic window will open.

YouTube Video:

Thanks...

Cadence Virtuoso

noreply@blogger.com (TechinGrow) — Thu, 27 Jan 2022 11:35:00 +0000

Cadence Virtuoso

Cadence Virtuoso is a software used to make electronic circuits. This software helps in doing various analysis of circuit. Cadence Virtuoso Software is basically used to make electronic circuits for fabrication process.

The Cadence Virtuoso System Design Platform links two world-class Cadence technologies—custom IC design and package/PCB design/analysis—creating a holistic methodology that automates and streamlines the design and verification flow for multi-die heterogeneous systems.

The Virtuoso System Design Platform allows IC designers to easily include system-level layout parasitics in the IC verification flow, enabling time savings by combining package/board layout connectivity data with the IC layout parasitic electrical model. The automatically generated “system-aware” schematic that results can then be easily used to create a testbench for final circuit-level simulation. The Virtuoso System Design Platform automates this entire flow, eliminating the highly manual and error-prone process of integrating system-level layout parasitic models back into the IC designer’s flow.

Above Image shows the analysis that you can do using Cadence Virtuoso. You can also do other analysis like Paramatric Analysis, also can use calculator to check graphical representation manually.

Hope so, that all of you are reading this Blog will understand the need of Cadence Virtuoso.

Thanks....

Service Lane || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:34:00 +0000

A driver is driving on the freeway. The check engine light of his vehicle is on, and the driver wants to get service immediately. Luckily, a service lane runs parallel to the highway. It varies in width along its length.

You will be given an array of widths at points along the road (indices), then a list of the indices of entry and exit points. Considering each entry and exit point pair, calculate the maximum size vehicle that can travel that segment of the service lane safely.

Example

If the entry index, and the exit, , there are two segment widths of and respectively. The widest vehicle that can fit through both is . If and , the widths are which limits vehicle width to .

Function Description

Complete the serviceLane function in the editor below.

serviceLane has the following parameter(s):

int n: the size of the array
int cases[t][2]: each element contains the starting and ending indices for a segment to consider, inclusive

Returns

int[t]: the maximum width vehicle that can pass through each segment of the service lane described

Input Format

The first line of input contains two integers, and , where denotes the number of width measurements and , the number of test cases. The next line has space-separated integers which represent the array .

The next lines contain two integers, and , where is the start index and is the end index of the segment to check.

Constraints

Sample Input

STDIN               Function
-----               --------
8 5                 n = 8, t = 5
2 3 1 2 3 2 3 3     width = [2, 3, 1, 2, 3, 2, 3, 3]
0 3                 cases = [[0, 3], [4, 6], [6, 7], [3, 5], [0, 7]]
4 6
6 7
3 5
0 7

Sample Output

Explanation

Below is the representation of the lane:

   |HIGHWAY|Lane|    ->    Width

0: |       |--|            2
1: |       |---|           3
2: |       |-|             1
3: |       |--|            2
4: |       |---|           3
5: |       |--|            2
6: |       |---|           3
7: |       |---|           3

: From index through we have widths and . Nothing wider than can pass all segments.
: From index through we have widht and . Nothing wider than can pass all segments.
: .
:
: .

Solution:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() 
{
    int i,n,j,t,min;
    scanf("%d%d",&n,&t);
    int* arr=(int*)malloc(n*sizeof(int));
    int p,q;
    for(i=0;i<n;i++)
    {
        scanf("%d",arr+i);
    }
    min=4;
    for(i=0;i<t;i++)
    {
        scanf("%d%d",&p,&q);
        min=4;
        for(j=p;j<=q;j++)
        {
            if(arr[j]<min)
                min=arr[j];
        }
        printf("%d\n",min);
    }
    return 0;
}

Viral Advertising || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:32:00 +0000

HackerLand Enterprise is adopting a new viral advertising strategy. When they launch a new product, they advertise it to exactly people on social media.

On the first day, half of those people (i.e., ) like the advertisement and each shares it with of their friends. At the beginning of the second day, people receive the advertisement.

Each day, of the recipients like the advertisement and will share it with friends on the following day. Assuming nobody receives the advertisement twice, determine how many people have liked the ad by the end of a given day, beginning with launch day as day .

Example
.

Day Shared Liked Cumulative
1      5     2       2
2      6     3       5
3      9     4       9
4     12     6      15
5     18     9      24

The progression is shown above. The cumulative number of likes on the day is .

Function Description

Complete the viralAdvertising function in the editor below.

viralAdvertising has the following parameter(s):

int n: the day number to report

Returns

int: the cumulative likes at that day

Input Format

A single integer, , the day number.

Constraints

Sample Input

Sample Output

Explanation

This example is depicted in the following diagram:

people liked the advertisement on the first day, people liked the advertisement on the second day and people liked the advertisement on the third day, so the answer is .

Solution:

#include<stdio.h>
int main()
{
    int n,k=5,i,m;
    scanf("%d",&n);
    k=(k/2);
    m=k;
    for(i=1;i<n;i++)
    {
        m=m*3;
        m=m/2;
        k=k+m;
    }
    printf("%d",k);
}

The Hurdle Race || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:30:00 +0000

A video player plays a game in which the character competes in a hurdle race. Hurdles are of varying heights, and the characters have a maximum height they can jump. There is a magic potion they can take that will increase their maximum jump height by unit for each dose. How many doses of the potion must the character take to be able to jump all of the hurdles. If the character can already clear all of the hurdles, return .

Example

The character can jump unit high initially and must take doses of potion to be able to jump all of the hurdles.

Function Description

Complete the hurdleRace function in the editor below.

hurdleRace has the following parameter(s):

int k: the height the character can jump naturally
int height[n]: the heights of each hurdle

Returns

int: the minimum number of doses required, always or more

Input Format

The first line contains two space-separated integers and , the number of hurdles and the maximum height the character can jump naturally.
The second line contains space-separated integers where .

Constraints

Sample Input 0

5 4
1 6 3 5 2

Sample Output 0

Explanation 0

Dan's character can jump a maximum of units, but the tallest hurdle has a height of :

To be able to jump all the hurdles, Dan must drink doses.

Sample Input 1

5 7
2 5 4 5 2

Sample Output 1

Explanation 1

Dan's character can jump a maximum of units, which is enough to cross all the hurdles:

Because he can already jump all the hurdles, Dan needs to drink doses.

Solution:

#include<stdio.h>
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    int a[n],i,j;
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    j=a[0];
    for(i=1;i<n;i++)
    {
        if(j>=a[i])
        j=j;
        else
        j=a[i];
    }
    if(j>m)
    printf("%d",j-m);
    else
    printf("0");
}

Cats and a Mouse || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:28:00 +0000

Two cats and a mouse are at various positions on a line. You will be given their starting positions. Your task is to determine which cat will reach the mouse first, assuming the mouse does not move and the cats travel at equal speed. If the cats arrive at the same time, the mouse will be allowed to move and it will escape while they fight.

You are given queries in the form of , , and representing the respective positions for cats and , and for mouse . Complete the function to return the appropriate answer to each query, which will be printed on a new line.

If cat catches the mouse first, print Cat A.
If cat catches the mouse first, print Cat B.
If both cats reach the mouse at the same time, print Mouse C as the two cats fight and mouse escapes.

Example

The cats are at positions (Cat A) and (Cat B), and the mouse is at position . Cat B, at position will arrive first since it is only unit away while the other is units away. Return 'Cat B'.

Function Description

Complete the catAndMouse function in the editor below.

catAndMouse has the following parameter(s):

int x: Cat 's position
int y: Cat 's position
int z: Mouse 's position

Returns

string: Either 'Cat A', 'Cat B', or 'Mouse C'

Input Format

The first line contains a single integer, , denoting the number of queries.
Each of the subsequent lines contains three space-separated integers describing the respective values of (cat 's location), (cat 's location), and (mouse 's location).

Constraints

Sample Input 0

2
1 2 3
1 3 2

Sample Output 0

Cat B
Mouse C

Explanation 0

Query 0: The positions of the cats and mouse are shown below:

Cat will catch the mouse first, so we print Cat B on a new line.

Query 1: In this query, cats and reach mouse at the exact same time:

Because the mouse escapes, we print Mouse C on a new line.

Solution:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n ; i++)
    {
        int x; 
        int y; 
        int z; 
        scanf("%d %d %d",&x,&y,&z);
        int a = (x - z);
        if(a<0)
        a=(-1)*a;
        int b = (y - z);
        if(b<0)
        b=(-1)*b;
        if (a < b) printf("Cat A\n");
        else if (b < a) printf("Cat B\n");
        else printf("Mouse C\n");
    }
}

Bill Division || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:26:00 +0000

Two friends Anna and Brian, are deciding how to split the bill at a dinner. Each will only pay for the items they consume. Brian gets the check and calculates Anna's portion. You must determine if his calculation is correct.

For example, assume the bill has the following prices: . Anna declines to eat item which costs . If Brian calculates the bill correctly, Anna will pay . If he includes the cost of , he will calculate . In the second case, he should refund to Anna.

Function Description

Complete the bonAppetit function in the editor below. It should print Bon Appetit if the bill is fairly split. Otherwise, it should print the integer amount of money that Brian owes Anna.

bonAppetit has the following parameter(s):

bill: an array of integers representing the cost of each item ordered
k: an integer representing the zero-based index of the item Anna doesn't eat
b: the amount of money that Anna contributed to the bill

Input Format

The first line contains two space-separated integers and , the number of items ordered and the -based index of the item that Anna did not eat.
The second line contains space-separated integers where .
The third line contains an integer, , the amount of money that Brian charged Anna for her share of the bill.

Constraints

The amount of money due Anna will always be an integer

Output Format

If Brian did not overcharge Anna, print Bon Appetit on a new line; otherwise, print the difference (i.e., ) that Brian must refund to Anna. This will always be an integer.

Sample Input 0

4 1
3 10 2 9
12

Sample Output 0

Explanation 0
Anna didn't eat item , but she shared the rest of the items with Brian. The total cost of the shared items is and, split in half, the cost per person is . Brian charged her for her portion of the bill. We print the amount Anna was overcharged, , on a new line.

Sample Input 1

4 1
3 10 2 9
7

Sample Output 1

Bon Appetit

Explanation 1
Anna didn't eat item , but she shared the rest of the items with Brian. The total cost of the shared items is and, split in half, the cost per person is . Because , we print Bon Appetit on a new line.

Solution:

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

int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    int a[n];
    int z;
    int sum=0;
    for(int i=0;i<n;i++){
        scanf("%d",a+i);
        if(i==k)
            z=a[i];
        else
            sum+=a[i];
    }
    int p=sum/2;
    int c;
    scanf("%d",&c);
    if(p==c)
        printf("Bon Appetit\n");
    else
        printf("%d\n",c-p);
    return 0;
}

Day of the Programmer || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:21:00 +0000

Marie invented a Time Machine and wants to test it by time-traveling to visit Russia on the Day of the Programmer (the 256th day of the year) during a year in the inclusive range from 1700 to 2700.

From 1700 to 1917, Russia's official calendar was the Julian calendar; since 1919 they used the Gregorian calendar system. The transition from the Julian to Gregorian calendar system occurred in 1918, when the next day after January 31st was February 14th. This means that in 1918, February 14th was the 32nd day of the year in Russia.

In both calendar systems, February is the only month with a variable amount of days; it has 29 days during a leap year, and 28 days during all other years. In the Julian calendar, leap years are divisible by 4; in the Gregorian calendar, leap years are either of the following:

Divisible by 400.
Divisible by 4 and not divisible by 100.

Given a year, , find the date of the 256th day of that year according to the official Russian calendar during that year. Then print it in the format dd.mm.yyyy, where dd is the two-digit day, mm is the two-digit month, and yyyy is .

For example, the given = 1984. 1984 is divisible by 4, so it is a leap year. The 256th day of a leap year after 1918 is September 12, so the answer is .

Function Description

Complete the dayOfProgrammer function in the editor below. It should return a string representing the date of the 256th day of the year given.

dayOfProgrammer has the following parameter(s):

year: an integer

Input Format

A single integer denoting year .

Constraints

1700 \le y \le 2700

Output Format

Print the full date of Day of the Programmer during year in the format dd.mm.yyyy, where dd is the two-digit day, mm is the two-digit month, and yyyy is .

Sample Input 0

Sample Output 0

13.09.2017

Explanation 0

In the year = 2017, January has 31 days, February has 28 days, March has 31 days, April has 30 days, May has 31 days, June has 30 days, July has 31 days, and August has 31 days. When we sum the total number of days in the first eight months, we get 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 = 243. Day of the Programmer is the 256th day, so then calculate 256 - 243 = 13 to determine that it falls on day 13 of the 9th month (September). We then print the full date in the specified format, which is 13.09.2017.

Sample Input 1

Sample Output 1

12.09.2016

Explanation 1

Year = 2016 is a leap year, so February has 29 days but all the other months have the same number of days as in 2017. When we sum the total number of days in the first eight months, we get 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 = 244. Day of the Programmer is the 256th day, so then calculate 256 - 244 = 12 to determine that it falls on day 12 of the 9th month (September). We then print the full date in the specified format, which is 12.09.2016.

Sample Input 2

Sample Output 2

12.09.1800

Explanation 2

Since 1800 is leap year as per Julian calendar. Day lies on 12 September.
Solution:

#include <stdio.h>

int main()
{
    int y,leap; 
    scanf("%d",&y);
    leap=0;
    if (y % 4 == 0)
    leap=1;
    if (y > 1918)
    {
    if (y % 100 == 0) leap=0;
    if (y % 400 == 0) leap=1;
    }
    if (y != 1918) 
    {
        if (leap == 0)
        printf("13.09.%d\n",y);
        else printf("12.09.%d\n",y);
    }
    else printf("26.09.%d\n",y);
    return 0;
}

Divisible Sum Pairs || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:12:00 +0000

Given an array of integers and a positive integer , determine the number of pairs where and + is divisible by .

Example

Three pairs meet the criteria: and .

Function Description

Complete the divisibleSumPairs function in the editor below.

divisibleSumPairs has the following parameter(s):

int n: the length of array
int ar[n]: an array of integers
int k: the integer divisor

Returns
- int: the number of pairs

Input Format

The first line contains space-separated integers, and .
The second line contains space-separated integers, each a value of .

Constraints

Sample Input

STDIN           Function
-----           --------
6 3             n = 6, k = 3
1 3 2 6 1 2     ar = [1, 3, 2, 6, 1, 2]

Sample Output

Solution:

#include<stdio.h>
int main()
{
    int n,i,j,k,m,c=0;
    scanf("%d",&n);
    scanf("%d",&k);
    int a[n];
    for(i=0;i<n;i++)
    scanf("%d",&a[i]);
    for(i=0;i<n;i++)
    {
        m=a[i];
        for(j=i+1;j<n;j++)
        {
            m+=a[j];
            if(m%k==0)
            c++;
            m=a[i];
        }
    }
    printf("%d",c);
    return 0;
}

Breaking The Records || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 10:04:00 +0000

Maria plays college basketball and wants to go pro. Each season she maintains a record of her play. She tabulates the number of times she breaks her season record for most points and least points in a game. Points scored in the first game establish her record for the season, and she begins counting from there.

Example

Scores are in the same order as the games played. She tabulates her results as follows:

                                     Count
    Game  Score  Minimum  Maximum   Min Max
     0           12              12       12              0   0
     1           24              12       24              0   1
     2           10              10       24              1   1
     3           24              10       24              1   1

Given the scores for a season, determine the number of times Maria breaks her records for most and least points scored during the season.

Function Description

Complete the breakingRecords function in the editor below.

breakingRecords has the following parameter(s):

int scores[n]: points scored per game

Returns

int[2]: An array with the numbers of times she broke her records. Index is for breaking most points records, and index is for breaking least points records.

Input Format

The first line contains an integer , the number of games.
The second line contains space-separated integers describing the respective values of .

Constraints

Sample Input 0

9
10 5 20 20 4 5 2 25 1

Sample Output 0

2 4

Explanation 0

The diagram below depicts the number of times Maria broke her best and worst records throughout the season:

She broke her best record twice (after games and ) and her worst record four times (after games , , , and ), so we print 2 4 as our answer. Note that she did not break her record for best score during game , as her score during that game was not strictly greater than her best record at the time.

Sample Input 1

10
3 4 21 36 10 28 35 5 24 42

Sample Output 1

4 0

Explanation 1

The diagram below depicts the number of times Maria broke her best and worst records throughout the season:

She broke her best record four times (after games , , , and ) and her worst record zero times (no score during the season was lower than the one she earned during her first game), so we print 4 0 as our answer.

Solution:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    int a[n],i,j,m,c=0,p=0;
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    m=a[0];
    for(i=1;i<n;i++)
    {
        if(m>=a[i])
        j=0;
        else
        {
        m=a[i];
        c++;
        }
    }
    m=a[0];
    for(i=1;i<n;i++)
    {
        if(m<=a[i])
        j=0;
        else
        {
        m=a[i];
        p++;
        }
    }
    printf("%d %d",c,p);
}

Apple and Orange || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:59:00 +0000

Sam's house has an apple tree and an orange tree that yield an abundance of fruit. Using the information given below, determine the number of apples and oranges that land on Sam's house.

In the diagram below:

The red region denotes the house, where is the start point, and is the endpoint. The apple tree is to the left of the house, and the orange tree is to its right.
Assume the trees are located on a single point, where the apple tree is at point , and the orange tree is at point .
When a fruit falls from its tree, it lands units of distance from its tree of origin along the -axis. *A negative value of means the fruit fell units to the tree's left, and a positive value of means it falls units to the tree's right. *

Given the value of for apples and oranges, determine how many apples and oranges will fall on Sam's house (i.e., in the inclusive range )?

For example, Sam's house is between and . The apple tree is located at and the orange at . There are apples and oranges. Apples are thrown units distance from , and units distance. Adding each apple distance to the position of the tree, they land at . Oranges land at . One apple and two oranges land in the inclusive range so we print

1
2

Function Description

Complete the countApplesAndOranges function in the editor below. It should print the number of apples and oranges that land on Sam's house, each on a separate line.

countApplesAndOranges has the following parameter(s):

s: integer, starting point of Sam's house location.
t: integer, ending location of Sam's house location.
a: integer, location of the Apple tree.
b: integer, location of the Orange tree.
apples: integer array, distances at which each apple falls from the tree.
oranges: integer array, distances at which each orange falls from the tree.

Input Format

The first line contains two space-separated integers denoting the respective values of  and .
The second line contains two space-separated integers denoting the respective values of  and .
The third line contains two space-separated integers denoting the respective values of  and .
The fourth line contains  space-separated integers denoting the respective distances that each apple falls from point .
The fifth line contains  space-separated integers denoting the respective distances that each orange falls from point .

Constraints

Output Format

Print two integers on two different lines:

The first integer: the number of apples that fall on Sam's house.
The second integer: the number of oranges that fall on Sam's house.

Sample Input 0

Sample Output 0

1
1

Explanation 0

The first apple falls at position .
The second apple falls at position .
The third apple falls at position .
The first orange falls at position .
The second orange falls at position .
Only one fruit (the second apple) falls within the region between and , so we print as our first line of output.
Only the second orange falls within the region between and , so we print as our second line of output.

Solution:

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main()
{
    int s; 
    int t; 
    scanf("%d %d",&s,&t);
    int a; 
    int b; 
    scanf("%d %d",&a,&b);
    int m; 
    int n; 
    int ans1=0,ans2=0;
    scanf("%d %d",&m,&n);
    int *apple = malloc(sizeof(int) * m);
    for(int apple_i = 0; apple_i < m; apple_i++){
       scanf("%d",&apple[apple_i]);
        if((apple[apple_i]+a)<=t && (apple[apple_i]+a)>=s)
            ans1++;
    }
    int *orange = malloc(sizeof(int) * n);
    for(int orange_i = 0; orange_i < n; orange_i++){
       scanf("%d",&orange[orange_i]);
        if((orange[orange_i]+b)<=t && (orange[orange_i]+b)>=s)
            ans2++;
    }
    printf("%d\n%d",ans1,ans2);
    return 0;
}

Grading Students || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:56:00 +0000

HackerLand University has the following grading policy:

Every student receives a in the inclusive range from to .
Any less than is a failing grade.

Sam is a professor at the university and likes to round each student's according to these rules:

If the difference between the and the next multiple of is less than , round up to the next multiple of .
If the value of is less than , no rounding occurs as the result will still be a failing grade.

Examples

round to (85 - 84 is less than 3)
do not round (result is less than 40)
do not round (60 - 57 is 3 or higher)

Given the initial value of for each of Sam's students, write code to automate the rounding process.

Function Description

Complete the function gradingStudents in the editor below.

gradingStudents has the following parameter(s):

int grades[n]: the grades before rounding

Returns

int[n]: the grades after rounding as appropriate

Input Format

The first line contains a single integer, , the number of students.
Each line of the subsequent lines contains a single integer, .

Constraints

Sample Input 0

Sample Output 0

Explanation 0

Student received a , and the next multiple of from is . Since , the student's grade is rounded to .
Student received a , and the next multiple of from is . Since , the grade will not be modified and the student's final grade is .
Student received a , and the next multiple of from is . Since , the student's grade will be rounded to .
Student received a grade below , so the grade will not be modified and the student's final grade is .

Solution:

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main()
{
    int n;
    scanf("%d",&n);
    for(int a0 = 0; a0 < n; a0++)
    {
        int grade;
        scanf("%d",&grade);
        int x = (grade+4)/5;
        x *= 5;
        if(x>=40 && x-grade<3)
        grade=x;
        printf("%d\n",grade);
    }
return 0;
}

Birthday Cake Candles || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:51:00 +0000

You are in charge of the cake for a child's birthday. You have decided the cake will have one candle for each year of their total age. They will only be able to blow out the tallest of the candles. Count how many candles are tallest.

Example

The maximum height candles are units high. There are of them, so return .

Function Description

Complete the function birthdayCakeCandles in the editor below.

birthdayCakeCandles has the following parameter(s):

int candles[n]: the candle heights

Returns

int: the number of candles that are tallest

Input Format

The first line contains a single integer, , the size of .
The second line contains space-separated integers, where each integer describes the height of .

Constraints

Sample Input 0

4
3 2 1 3

Sample Output 0

Explanation 0

Candle heights are . The tallest candles are units, and there are of them.

Solution:

#include <stdio.h>
int main()
{
    int i;
    int n;
    int max=0;
    scanf("%d",&n);
    int a[n];
    int count=0;
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if (max<a[i])
        max=a[i];
    }
for(i=1;i<=n;i++)
if(a[i]==max)
count++;
printf("%d",count);
return 0;
}

Min-Max Sum || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:23:00 +0000

Given five positive integers, find the minimum and maximum values that can be calculated by summing exactly four of the five integers. Then print the respective minimum and maximum values as a single line of two space-separated long integers.

Example

The minimum sum is and the maximum sum is . The function prints

16 24

Function Description

Complete the miniMaxSum function in the editor below.

miniMaxSum has the following parameter(s):

arr: an array of integers

Print two space-separated integers on one line: the minimum sum and the maximum sum of of elements.

Input Format

A single line of five space-separated integers.

Constraints

Output Format

Print two space-separated long integers denoting the respective minimum and maximum values that can be calculated by summing exactly four of the five integers. (The output can be greater than a 32 bit integer.)

Sample Input

1 2 3 4 5

Sample Output

10 14

Explanation

The numbers are , , , , and . Calculate the following sums using four of the five integers:

Sum everything except , the sum is .
Sum everything except , the sum is .
Sum everything except , the sum is .
Sum everything except , the sum is .
Sum everything except , the sum is .

Hints: Beware of integer overflow! Use 64-bit Integer.

Need help to get started? Try the Solve Me First problem

Solution:

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

int main()
{    
    int a[5];
    long s=0;
    for(int i=0;i<5;i++)
    {
        scanf("%d",a+i);
        s+=a[i];
    }
    int min=a[0];
    int max=a[0];
    for(int i=1;i<5;i++)
    {
        if(a[i]>max)
            max=a[i];
        if(a[i]<min)
            min=a[i];
    }
    printf("%ld %ld",s-max,s-min);
    return 0;
}

Staircase || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:17:00 +0000

Staircase detail

This is a staircase of size :

   #
  ##
 ###
####

Its base and height are both equal to . It is drawn using # symbols and spaces. The last line is not preceded by any spaces.

Write a program that prints a staircase of size .

Function Description

Complete the staircase function in the editor below.

staircase has the following parameter(s):

int n: an integer

Print a staircase as described above.

Input Format

A single integer, , denoting the size of the staircase.

Constraints

Output Format

Print a staircase of size using # symbols and spaces.

Note: The last line must have spaces in it.

Sample Input

Sample Output

     #
    ##
   ###
  ####
 #####
######

Explanation

The staircase is right-aligned, composed of # symbols and spaces, and has a height and width of .

Solution:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    int i,j,k;
    char a[n],c='#',h=' ';
    for(i=n-1;i>=0;i--)
    {
        for(j=0;j<n;j++)
        {
            if(j>=i)
            a[j]=c;
            else
            a[j]=h;
        }
        for(k=0;k<n;k++)
        {
        printf("%c",a[k]);
        }
        printf("\n");
    }
}

Plus Minus || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:15:00 +0000

Given an array of integers, calculate the ratios of its elements that are positive, negative, and zero. Print the decimal value of each fraction on a new line with places after the decimal.

Note: This challenge introduces precision problems. The test cases are scaled to six decimal places, though answers with absolute error of up to are acceptable.

Example

There are elements, two positive, two negative and one zero. Their ratios are , and . Results are printed as:

0.400000
0.400000
0.200000

Function Description

Complete the plusMinus function in the editor below.

plusMinus has the following parameter(s):

int arr[n]: an array of integers

Print
Print the ratios of positive, negative and zero values in the array. Each value should be printed on a separate line with digits after the decimal. The function should not return a value.

Input Format

The first line contains an integer, , the size of the array.
The second line contains space-separated integers that describe .

Constraints

Output Format

Print the following lines, each to decimals:

proportion of positive values
proportion of negative values
proportion of zeros

Sample Input

STDIN           Function
-----           --------
6               arr[] size n = 6
-4 3 -9 0 4 1   arr = [-4, 3, -9, 0, 4, 1]

Sample Output

0.500000
0.333333
0.166667

Explanation

There are positive numbers, negative numbers, and zero in the array.
The proportions of occurrence are positive: , negative: and zeros: .

Solution:

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

int main()
{
    int n,a[100],i;
    float m=0,z=0,p=0;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]>0)
        {
            p++;    
        }
        else if(a[i]==0)
        {
            z++;  
        }
        else
        {
            m++;    
        }
    }
    printf("%.3f\n%.3f\n%.3f\n",p/n,m/n,z/n);   
    return 0;
}

Diagonal Difference || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:08:00 +0000

Given a square matrix, calculate the absolute difference between the sums of its diagonals.

For example, the square matrix is shown below:

1 2 3
4 5 6
9 8 9

The left-to-right diagonal = . The right to left diagonal = . Their absolute difference is .

Function description

Complete the function in the editor below.

diagonalDifference takes the following parameter:

int arr[n][m]: an array of integers

Return

int: the absolute diagonal difference

Input Format

The first line contains a single integer, , the number of rows and columns in the square matrix .
Each of the next lines describes a row, , and consists of space-separated integers .

Constraints

Output Format

Return the absolute difference between the sums of the matrix's two diagonals as a single integer.

Sample Input

3
11     2       4
4         5       6
10     8      -12

Sample Output

Explanation

The primary diagonal is:

11
         5
                -12

Sum across the primary diagonal: 11 + 5 - 12 = 4

The secondary diagonal is:

               4
         5
10

Sum across the secondary diagonal: 4 + 5 + 10 = 19
Difference: |4 - 19| = 15

Note: |x| is the absolute value of x

Solution:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    int i,j,a[n][n],k=0,s=0;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i==j)
            {
                k=k+a[i][j];
            }
            if((i+j)==(n-1))
            {
                s=s+a[i][j];
            }
        }
    }
    int m;
    m=k-s;
    if(m<0)
    printf("%d",(-1)*m);
    else if(m>=0)
    printf("%d",m);
}

A Very Big Sum || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:04:00 +0000

In this challenge, you are required to calculate and print the sum of the elements in an array, keeping in mind that some of those integers may be quite large.

Function Description

Complete the aVeryBigSum function in the editor below. It must return the sum of all array elements.

aVeryBigSum has the following parameter(s):

int ar[n]: an array of integers .

Return

long: the sum of all array elements

Input Format

The first line of the input consists of an integer .
The next line contains space-separated integers contained in the array.

Output Format

Return the integer sum of the elements in the array.

Constraints

Sample Input

1000000001 1000000002 1000000003 1000000004 1000000005

Output

5000000015

Note:

The range of the 32-bit integer is .

When we add several integer values, the resulting sum might exceed the above range. You might need to use long int C/C++/Java to store such sums.

Solution:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    long int a[n]; 
    for(int i=0;i<n;i++)
    {
        scanf("%ld",&a[i]);
    }
    long int s=0;
    for(int i=0;i<n;i++)
    {
        s=s+a[i];
    }
    printf("%ld",s);
}

Simple Array Sum || HackerRank C Programming Solution

noreply@blogger.com (TechinGrow) — Mon, 01 Nov 2021 09:01:00 +0000

Given an array of integers, find the sum of its elements.

For example, if the array , , so return .

Function Description

Complete the simpleArraySum function in the editor below. It must return the sum of the array elements as an integer.

simpleArraySum has the following parameter(s):

ar: an array of integers

Input Format

The first line contains an integer, , denoting the size of the array.
The second line contains space-separated integers representing the array's elements.

Constraints

Output Format

Print the sum of the array's elements as a single integer.

Sample Input

1 2 3 4 10 11

Sample Output

Explanation

We print the sum of the array's elements: .

Solution:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
    int n;
    scanf("%d",&n);
    int a[n];
    long int s=0;
    for(int i = 0; i<n; i++)
      scanf("%d",&a[i]);
    for(int i=0; i<n;i++)
        s+= a[i]; 
    printf("%ld",s);
    return 0;
}