CODES

PERL SCRIPT for killing CLOSE_WAIT

2015-11-17T18:13:00.000+05:30

Some time our application gets chocked because of large number of connection in COSE WAIT state, In such case there is only way to get rid of it by closing application, but in production we cann't kill our application this frequently, so here is a hack I come accross for it.

Ultimate solution :: optimise and find reason in your application.

#!/bin/bash
netstat -tulnap | grep CLOSE_WAIT | awk '{print $4,$5}' | sed 's/:/ /g' | awk '{print $2, $3, $5, $6}'| while read line; do
echo $line
perl kill_cw.pl $line;
done

wget http://apt.sw.be/redhat/el6/en/x86_64/rpmforge/RPMS/perl-Net-RawIP-0.25-1.el6.rf.x86_64.rpm
wget http://apt.sw.be/redhat/el6/en/x86_64/rpmforge/RPMS/perl-Net-Pcap-0.16-1.el6.rf.x86_64.rpm
wget http://www6.atomicorp.com/channels/atomic/centos/6/i386/RPMS/perl-NetPacket-0.41.1-1.el6.art.noarch.rpm

rpm -Uvh perl-Net-RawIP-0.25-1.el6.rf.x86_64.rpm
rpm -Uvh perl-Net-Pcap-0.16-1.el6.rf.x86_64.rpm
yum install perl-Net-Pcap

rpm -Uvh perl-NetPacket-0.41.1-1.el6.art.noarch.rpm

(http://pkgs.org/centos-6/atomic-i386/perl-NetPacket-0.41.1-1.el6.art.noarch.rpm.html)

Install PhantomJS on Ubuntu

2013-12-20T20:30:00.001+05:30


cd /usr/local/share

sudo wget http://phantomjs.googlecode.com/files/phantomjs-1.9.2-linux-x86_64.tar.bz2

sudo tar xjf phantomjs-1.9.2-linux-x86_64.tar.bz2

sudo ln -s /usr/local/share/phantomjs-1.9.2-linux-x86_64/bin/phantomjs /usr/local/share/phantomjs

sudo ln -s /usr/local/share/phantomjs-1.9.2-linux-x86_64/bin/phantomjs /usr/local/bin/phantomjs

sudo ln -s /usr/local/share/phantomjs-1.9.2-linux-x86_64/bin/phantomjs /usr/bin/phantomjs

on this,

phantomjs -v

it will give
1.9.2

Some learning on Cassandra - I

2013-10-22T15:04:00.002+05:30

In this post we will not talk about Cassandra – Data Model, There are many blogs on them and one of the best describe is by Animesh Kumar, oh yup Animesh sir is my collage(ISM) senior.

poor approach:

Serialize the Object into JSON.
Now store this JSON as a string into Cassandra.
When need to get the object, read JSON string, deserialize the JSON.

The above approach is fine and works well, but It have its own disadvantage. Say we have
Object(as JSON):
{
name: "nijju"
country: "IN"
email: "niraj.nijju@gmail.com"
}

and we store it into a column into cassandra( as string).
When we need to update a field say country then we need to read the whole JSON string, then we need to deserialize it, and then to change the object attribute and again serialize the resulting object into JSON and storing this JSON at that key.

As in cassandra write is much faster than read operation, so making a read before write will not be a good approach.
Even if your application is mostly write-once data then for reading a single field will make to read all.

In conclusion, we can better performance by storing all the fields into a separate column. For making an update on a field we only need to re-write the corresponding column.

It totally depends on our read behavior, as rows will be store on random disk, so reading from various rows at same time will be slow.
and If rows will be too long then these columns will be stored together on same node so we may have scalability challenges.

Let me know your thoughts

Python script for get the files_from_directory, using_grep_on_file, on_day_increment

2013-09-28T20:00:00.000+05:30

How can I find all files in directory with the extension .txt in python?

http://stackoverflow.com/questions/3964681/find-all-files-in-directory-with-extension-txt-with-python

import glob
import os
#os.chdir("/mydir")
for files in glob.glob("*.txt"):
    print files

python grep on file

for line in open("file_name"):
 if "search_string" in line:
   print line

Python increment day in datetime:

import datetime
today = datetime.datetime.today()
today += datetime.timedelta(days=1)

Google Code Jam Problem C. Fair and Square

2013-04-14T15:47:00.000+05:30

Below is my solution for Problem C. Fair and Square

Well I qualified for Google Code Jam 2013 with only 60 points.

* Still to add method to handle Large Input 2.

Input is taken from a file(C-small-practice.in)

Method 1:

* @author niraj.nijju

package gcj.TwelveThirteen;

import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.InputStreamReader;

public class SmallOfFairAndSquare {

public static void main(String [] args){
try{
// System.out.println(Double.MAX_VALUE);
// System.exit(0);

  FileInputStream fstream = new FileInputStream("C-small-practice.in");
  DataInputStream in = new DataInputStream(fstream);
  BufferedReader br = new BufferedReader(new InputStreamReader(in));
  String strLine;
  strLine = br.readLine();
  int T = Integer.valueOf(strLine);
  System.out.println("T: "+T);
  int i=1;
  while ((strLine = br.readLine()) != null) {
//   System.out.println (strLine);
  String [] str = strLine.split(" ");
  double x = Double.valueOf(str[0].trim() );
  double y = Double.valueOf(str[1].trim() );
  int count= getFairSqur(x,y);
  System.out.println ("Case #"+i++ +": "+count);
  }
  in.close();
}catch (Exception e){//Catch exception if any
  System.err.println("Error: " + e.getMessage());
  e.printStackTrace();
}
}

public static int getFairSqur(double x, double y){
int count=0;
long x1 = (long)Math.ceil(Math.sqrt(x));
long y1 = (long)Math.sqrt(y);

for(long i=x1; i<=y1; i++ ){
if(checkPalindrom(i)){
long sq=i*i;
if(checkPalindrom(sq)){
// System.out.println("("+i+")");
count++;
}
}
}

return count;
}

public static Boolean checkPalindrom(long n)
{
   String normal = String.valueOf(n);
   //reverse the letters in string
   int length = normal.length(); //length of string
   StringBuffer result = new StringBuffer(length);
   int i;
   for(i = length - 1; i>=0; i--)
   {
   result.append(normal.charAt(i));
   }
   long r = Long.valueOf(result.toString());

   if(r==n)
   return true;
   return false;
}

}

Method 2:

* @author niraj.nijju

package gcj.TwelveThirteen;

import java.io.BufferedReader;

import java.io.DataInputStream;

import java.io.FileInputStream;

import java.io.InputStreamReader;

import org.omg.CORBA.INTERNAL;

public class CopyOfFairAndSquare {

public static long [] array = new long[10000010];

public static void main(String [] args){

fillArray();

try{

// System.out.println(Double.MAX_VALUE);

// System.exit(0);

FileInputStream fstream = new FileInputStream("C-small-practice.in");

DataInputStream in = new DataInputStream(fstream);

BufferedReader br = new BufferedReader(new InputStreamReader(in));

String strLine;

strLine = br.readLine();

int T = Integer.valueOf(strLine);

// System.out.println("T: "+T);

int i=1;

while ((strLine = br.readLine()) != null) {

// System.out.println (strLine);

String [] str = strLine.split(" ");

double x = Double.valueOf(str[0].trim() );

double y = Double.valueOf(str[1].trim() );

//int count= getFairSqur(x,y);

long count=0;

int xr = (int)Math.sqrt(x);

int yr = (int)Math.sqrt(y);

count = array[yr]-array[xr];

if(xr*xr == (int)x){

if(checkPalindrom(xr) && checkPalindrom(xr*xr)){

count++;

}

System.out.println ("Case #"+i++ +": "+count);

}

in.close();

}catch (Exception e){//Catch exception if any

System.err.println("Error: " + e.getMessage());

e.printStackTrace();

}

public static void fillArray(){

long count = 0;

for(int i=1;i<array.length;i++){

long j = (long)i;

if(!checkPalindrom(j))

array[i]= count;

else if(!checkPalindrom(j*j))

array[i]= count;

else{

count++;

array[i]= count;

}

public static Boolean checkPalindrom(long n)

{

String normal = String.valueOf(n);

//reverse the letters in string

int length = normal.length(); //length of string

StringBuffer result = new StringBuffer(length);

int i;

for(i = length - 1; i>=0; i--)

{

result.append(normal.charAt(i));

}

long r=0;

try{

r= Long.valueOf(result.toString());

}catch(Exception e){

e.printStackTrace();

System.out.println("i:"+i+"\tr:"+r+"\tn:"+n);

}

if(r==n)

return true;

return false;

}

Google Code Jam Problem B. Lawnmower

2013-04-14T15:27:00.000+05:30

Below is my solution for Problem B: Lawnmower

Well I qualified for Google Code Jam 2013 with only 60 points.

Input is taken from a file(B-small-practice.in)

/*
* @author niraj.nijju
*/

package gcj.TwelveThirteen;

import java.io.BufferedReader;

import java.io.DataInputStream;

import java.io.FileInputStream;

import java.io.InputStreamReader;

public class Lawnmower {

public static void main(String [] args){

try{

FileInputStream fstream = new FileInputStream("B-small-practice.in");

DataInputStream in = new DataInputStream(fstream);

BufferedReader br = new BufferedReader(new InputStreamReader(in));

String strLine;

strLine = br.readLine();

int T = Integer.valueOf(strLine);

// System.out.println("T: "+T);

int i=1;

while(T-- > 0){

strLine = br.readLine();

String [] nm = strLine.split(" ");

int n = Integer.valueOf(nm[0].trim());

int m = Integer.valueOf(nm[1].trim());

String [][] lawn = new String[n+1][m+1];

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

String ms = br.readLine();

ms+= " 0";

lawn[j]=ms.split(" ");

}

if(canDo(lawn,n,m) ){

System.out.println ("Case #"+i++ +": YES");

}else{

System.out.println ("Case #"+i++ +": NO");

}

in.close();

}catch (Exception e){//Catch exception if any

System.err.println("Error: " + e.getMessage());

e.printStackTrace();

}

public static Boolean canDo(String [][] lawn, int n, int m){

getLargest(lawn,n,m);

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

for(int j=0;j<m;j++){

if(Integer.valueOf(lawn[i][j])<Integer.valueOf(lawn[i][m])

&& Integer.valueOf(lawn[i][j]) <Integer.valueOf(lawn[n][j]) )

return false;

}

return true;

}

public static void getLargest(String [][] lawn,int n, int m){

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

int hMax=0;

for(int j=0;j<m;j++){

int val=Integer.valueOf(lawn[i][j]);

if(hMax < val){

hMax = val;

}

lawn[i][m]=String.valueOf(hMax);

}

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

int vMax=0;

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

int val=Integer.valueOf(lawn[j][i]);

if(vMax < val){

vMax = val;

}

lawn[n][i]=String.valueOf(vMax);

}

Google Code Jam Problem A. Tic-Tac-Toe-Tomek

2013-04-14T15:18:00.000+05:30

Below is my solution for Problem A: Tic-Tac-Toe-Tomek

Well I qualified for Google Code Jam 2013 with only 60 points.

Problem

Tic-Tac-Toe-Tomek is a game played on a 4 x 4 square board. The board starts empty, except that a single 'T' symbol may appear in one of the 16 squares. There are two players: X and O. They take turns to make moves, with X starting. In each move a player puts her symbol in one of the empty squares. Player X's symbol is 'X', and player O's symbol is 'O'.

After a player's move, if there is a row, column or a diagonal containing 4 of that player's symbols, or containing 3 of her symbols and the 'T' symbol, she wins and the game ends. Otherwise the game continues with the other player's move. If all of the fields are filled with symbols and nobody won, the game ends in a draw. See the sample input for examples of various winning positions.

Given a 4 x 4 board description containing 'X', 'O', 'T' and '.' characters (where '.' represents an empty square), describing the current state of a game, determine the status of the Tic-Tac-Toe-Tomek game going on. The statuses to choose from are:

"X won" (the game is over, and X won)
"O won" (the game is over, and O won)
"Draw" (the game is over, and it ended in a draw)
"Game has not completed" (the game is not over yet)

If there are empty cells, and the game is not over, you should output "Game has not completed", even if the outcome of the game is inevitable.

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of 4 lines with 4 characters each, with each character being 'X', 'O', '.' or 'T' (quotes for clarity only). Each test case is followed by an empty line.

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is one of the statuses given above. Make sure to get the statuses exactly right. When you run your code on the sample input, it should create the sample output exactly, including the "Case #1: ", the capital letter "O" rather than the number "0", and so on.

Limits

The game board provided will represent a valid state that was reached through play of the game Tic-Tac-Toe-Tomek as described above.

Small dataset

1 ≤ T ≤ 10.

Large dataset

1 ≤ T ≤ 1000.

Sample

Input	Output
`6 XXXT .... OO.. .... XOXT XXOO OXOX XXOO XOX. OX.. .... .... OOXX OXXX OX.T O..O XXXO ..O. .O.. T... OXXX XO.. ..O. ...O`	`Case #1: X won Case #2: Draw Case #3: Game has not completed Case #4: O won Case #5: O won Case #6: O won`

Note

Although your browser might not render an empty line after the last test case in the sample input, in a real input file there would be one.

Input is taken from a file(A-small-practice.in)

/*
* @author niraj.nijju
*/

package gcj.TwelveThirteen;

import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class TicTacToeTomek {

public static void main(String [] args){
try{
FileInputStream fstream = new FileInputStream("A-small-practice.in");
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
String strLine;
strLine = br.readLine();
int T = Integer.valueOf(strLine);
System.out.println("T: "+T);
int i=1;
while ((strLine = br.readLine()) != null) {
List<String> string = new ArrayList<String>();
string.add(strLine);
string.add(br.readLine());
string.add(br.readLine());
string.add(br.readLine());
if (T-- >0){
br.readLine();
}
int st = status(string, 1);
if(st==1)
System.out.println ("Case #"+i++ +": X won");
if(st==2)
System.out.println ("Case #"+i++ +": O won");
if(st==0)
System.out.println ("Case #"+i++ +": Draw");
if(st==3)
System.out.println ("Case #"+i++ +": Game has not completed");

//System.out.println (strLine);
}
in.close();
}catch (Exception e){//Catch exception if any
System.err.println("Error: " + e.getMessage());
e.printStackTrace();
}
}

public static int status(List<String> string, int t){
int status=0; // (1.X 2.O 0.draw 3.incomplete)
int i=0;
Boolean ic=false;
List<String> sl = new ArrayList<String>();
sl.add("");sl.add("");
sl.add("");sl.add("");
for(i=0;i<4;i++){
String s = string.get(i);
if(s.contains("."))
ic = true;
for(int z=0;z<4;z++){
sl.set(z, sl.get(z)+s.charAt(z));
}

if(s.equalsIgnoreCase("XXXX"))
return 1;
else if(s.equalsIgnoreCase("OOOO"))
return 2;

if(s.contains("T")){
int xCount=0, oCount=0;
for(int j=0;j<4;j++){
if( s.charAt(j) == 'X')
xCount++;
else if( s.charAt(j) == 'O')
oCount++;
// else if( s.charAt(j) == '.')
// ic = true;
}
if(xCount==3)
return 1;
if(oCount==3)
return 2;
}

}
if( (string.get(0).charAt(0)=='X' || string.get(0).charAt(0)=='T' )
&& (string.get(1).charAt(1)=='X' ||string.get(1).charAt(1)=='T')
&& (string.get(2).charAt(2)=='X' ||string.get(2).charAt(2)=='T')
&& (string.get(3).charAt(3)=='X' ||string.get(3).charAt(3)=='T') )
return 1;
if( (string.get(0).charAt(0)=='O' || string.get(0).charAt(0)=='T' )
&& (string.get(1).charAt(1)=='O' ||string.get(1).charAt(1)=='T')
&& (string.get(2).charAt(2)=='O' ||string.get(2).charAt(2)=='T')
&& (string.get(3).charAt(3)=='O' ||string.get(3).charAt(3)=='T') )
return 2;

if( (string.get(3-0).charAt(0)=='X' || string.get(3-0).charAt(0)=='T' )
&& (string.get(3-1).charAt(1)=='X' ||string.get(3-1).charAt(1)=='T')
&& (string.get(3-2).charAt(2)=='X' ||string.get(3-2).charAt(2)=='T')
&& (string.get(3-3).charAt(3)=='X' ||string.get(3-3).charAt(3)=='T') )
return 1;
if( (string.get(3-0).charAt(0)=='O' || string.get(0).charAt(3-0)=='T' )
&& (string.get(3-1).charAt(1)=='O' ||string.get(3-1).charAt(1)=='T')
&& (string.get(3-2).charAt(2)=='O' ||string.get(3-2).charAt(2)=='T')
&& (string.get(3-3).charAt(3)=='O' ||string.get(3-3).charAt(3)=='T') )
return 2;

if(t==1){
return status(sl,2);

}

if(ic)
return 3;
return status;
}

}

write a function to check given string matches with given pattern

2012-11-25T20:05:00.000+05:30

public class isMatchingWithWildcard {

public static void main(String [] args){

String pattern = "*abc*def*.doc*";

String str = "adsfabcxyzdefgh.do1docx";

if(isMatching(str, pattern))

System.out.print("Matching");

else

System.out.print("Not Matching");

}

public static Boolean isMatching(String str, String pattern){

int l = pattern.length();

if(pattern.lastIndexOf("*") == l-1)

pattern= (String) pattern.subSequence(0, l-1);

if(pattern.charAt(0) == '*')

pattern= (String) pattern.subSequence(1, pattern.length());

pattern = pattern.replace("*", "__");

String [] patternArray = pattern.split("__");

String sorttenString =str;

for(String aPattern:patternArray){

String [] temp = sorttenString.split(aPattern);

if(temp.length==1 && (aPattern == patternArray[patternArray.length-1]

&& !sorttenString.toLowerCase().contains(aPattern.toLowerCase())))

return false;

str.substring(temp[0].length()+aPattern.length());

}

return true;

}

Let me know your thoughts

Finding Pythagorean triplets in an array.

2012-11-25T14:54:00.001+05:30

// WORKING code in java with time O(n^2)

import java.util.Arrays;

public class PythagoreanTripletsInArray {

public static double [] getSquare(String [] args){

double [] array = new double[args.length];

for(int i=0;i<args.length;i++){

double j=Integer.parseInt(args[i]);

array[i]=j*j;

}

return array;

}

public static void main(String [] args ){

double [] array =getSquare(args);

Arrays.sort(array);

Boolean flage = true;

for(double aInt : array){

for(int i=0,j=(array.length-1);i<j;){

double sum = array[i]+array[j];

if(aInt == sum){

System.out.println(Math.sqrt(aInt)+", "+Math.sqrt(array[i])+", "+Math.sqrt(array[j]));

flage = false;

break;

}

else if(aInt > sum)

i++;

else

j--;

}

if(flage)

System.out.print("No such solution exist");

}

Let me know your thoughts

Re-arrange the odd/even to odd/even places

2012-10-02T21:13:00.000+05:30

#include<stdio.h>

void Swap (int A[], int i, int j)
{
int temp = A[i];
A[i]=A[j];
A[j]=temp;
}

int main()
{
int A[] = {3,1,4,5,7,6,10,8};
int i=0,j=1,n;
n=sizeof(A)/sizeof(int);
while(i<n)
{
while(i<n && A[i]%2 == 0)
i+=2;
while(j<n && A[j]%2 == 1)
j+=2;
if(i<n && j<n)
Swap(A,i,j);
}// end of while

for(i=0;i<n;i++)
printf(" %d",A[i]);

return 0;
}

Let me know your thoughts

Sort an Array of 0 1& 2 in one pass

2012-09-14T19:54:00.000+05:30

#include<stdio.h>

int main()
{
int A[] = {2,2,2,1,1,1,0,0,0,1,2,0};
int j,i,x,y,n,t=A[0];
n=sizeof(A)/sizeof(A[0]);
printf("Size of array : %d\n",n);
for(x=i=0,y=j=n-1;i<=j;)
{
if(t==0)
{
A[x]=0;
x++;
i++;
t=A[i];
}
else if(t==2)
{
t=A[y];
A[y]=2;
y--;
j--;
}
else
{
i++;
t=A[i];
}
}

while(x<=y)
{
A[x++]=1;
}

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

Let me know your thoughts.

Given a 2–d matrix , which has only 1’s and 0’s in it. Find the total number of connected sets in that matrix.

2012-07-25T21:34:00.001+05:30

Explanation:

Connected set can be defined as group of cell(s) which has 1 mentioned on it and have at least one other cell in that set with which they share the neighbor relationship. A cell with 1 in it and no surrounding neighbor having 1 in it can be considered as a set with one cell in it. Neighbors can be defined as all the cells adjacent to the given cell in 8 possible directions ( i.e N , W , E , S , NE , NW , SE , SW direction ). A cell is not a neighbor of itself.

Input format :

First line of the input contains T , number of test-cases.

Then follow T testcases. Each testcase has given format.

N [ representing the dimension of the matrix N X N ].

Followed by N lines , with N numbers on each line.

Ouput format :

For each test case print one line , number of connected component it has.

Sample Input :

0 0 1 0

1 0 1 0

0 1 0 0

1 1 1 1

1 0 0 1

0 0 0 0

0 1 1 0

1 0 0 1

1 0 0 1 1

0 0 1 0 0

0 0 0 0 0

1 1 1 1 1

0 0 0 0 0

0 0 1 0 0 1 0 0

1 0 0 0 0 0 0 1

0 0 1 0 0 1 0 1

0 1 0 0 0 1 0 0

1 0 0 0 0 0 0 0

0 0 1 1 0 1 1 0

1 0 1 1 0 1 1 0

0 0 0 0 0 0 0 0

Sample output :

Constraint :

0 < T < 6

0 < N < 1009

=================WORKING CODE=================

#include<stdio.h>

int count_connected_set(int A[1015][1015],int i,int j,int n)

{

A[i][j]=2;

if(A[i-1][j-1]==1 && i>0 && j>0 && i<=n && j<=n ) //NW

count_connected_set(A,i-1,j-1,n);

if(A[i-1][j]==1 && i>0 && j>0 && i<=n && j<=n) //N

count_connected_set(A,i-1,j,n);

if(A[i-1][j+1]==1 && i>0 && j>0 && i<=n && j<=n)// //NE

count_connected_set(A,i-1,j+1,n);

if(A[i][j-1]==1 && i>0 && j>0 && i<=n && j<=n) //W

count_connected_set(A,i,j-1,n);

if(A[i][j+1]==1 && i>0 && j>0 && i<=n && j<=n) //E

count_connected_set(A,i,j+1,n);

if(A[i+1][j-1]==1 && i>0 && j>0 && i<=n && j<=n) //SW

count_connected_set(A,i+1,j-1,n);

if(A[i+1][j]==1 && i>0 && j>0 && i<=n && j<=n) //S

count_connected_set(A,i+1,j,n);

if(A[i+1][j+1]==1 && i>0 && j>0 && i<=n && j<=n) //SE

count_connected_set(A,i+1,j+1,n);

return 1;

}// end of count_connected_set()

int main()

{

int i,j,t,k,n,A[1015][1015],count;

scanf("%d",&t);

while(t--)

{

scanf("%d",&n);

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

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

scanf("%d",&A[i][j]); // geting input a test case;

count =0;

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

{

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

if(A[i][j]==1)

if(count_connected_set(A,i,j,n))

count++;

}// of for

printf("%d\n",count);

}// end of while

return 0;

}

Let me know your thoughts.

REVERSE a string and if any charecter is repeating then keep that one comes last

2012-07-14T18:56:00.002+05:30

#include<stdio.h>
#include<string.h>
int main()
{
char S[20],S1[20];
int i,x,j,l,A[26]={0},count=0;
gets(S); // UPPER CASE ONLY
l=strlen(S);
for(i=0;i<l;i++)
{
x=S[i]-'A'+1;
if(!A[x])
count++;
A[x]++;
}
S1[count]=NULL;
for(j=l-1,i=0;count>i;)
{
x=S[j]-'A'+1;
if(A[x]>0)
{
S1[i++]=S[j];
A[x]=0;
}
j--;
}
printf("\n> %s",S1);
return 0;
}

WAP to find the position of a given word in a given text

2012-07-10T11:20:00.000+05:30

#include<stdio.h>
#include<string.h>
int main()
{
char S[100],W[20];
int i,j,l,m,flage=0;
printf("Enter Text : ");
gets(S);
printf("\nEnter Word : ");
gets(W);
l=strlen(S);
m=strlen(W);
for(i=0;i<l;i++)
{
if(S[i]==W[0])
{
for(j=1;j<m;j++)
{
if(S[i+j]!=W[j])
break ;
}
if(j==m)
{
if(!flage)
printf("\nPosition : ");
flage =1;
printf(" %d",i);
//flage=
}
}
}
if(!flage)
printf("\nNot found \n");
return 0;
}

Let me know your thoughts.

Which type of triangle is ?

2012-07-09T11:46:00.000+05:30

#include<stdio.h>
int which_triangle(int a,int b,int c)
{
if( !(a&&b&&c&&(a+b)>c&&(a+c)>b&&(b+c)>a) )
return 4;
else if(a==b && a==c)
return 3;
else if(a==b || a==c || b==c)
return 2;
else
return 1;
}

int main()
{
int a,b,c,x;
printf("enter 3 side length : ");
scanf("%d %d %d",&a,&b,&c);
x=which_triangle(a,b,c);
switch(x){
case 1:
printf("\nScalene Triangle\n");
break ;
case 2:
printf("\nIsoscalene Triangle\n");
break ;
case 3:
printf("\nEquilateral Triangle\n");
break ;
case 4:
printf("\nError\n");
break ;
}// end of switch
return 0;
}

Let me know your thoughts.

Find the number of substrings of a string that are palindromes

2012-07-08T21:11:00.002+05:30

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

int is_palindrom(char S[20],int f,int l)
{
for(;f<=l;f++,l--)
{
if(S[f]!=S[l])
return 0;
}
return 1;
}
int main()
{
char S[20];
int i,l,j,k,count=0;
gets(S);
l=strlen(S);
for(i=0;i<l;i++)
{
for(j=i+1;j<l;j++)
{
if(S[i]==S[j])
if(is_palindrom(S,i,j))
{
printf("\n");
count++;
for(k=i;k<=j;k++)
printf("%c",S[k]);
}
}
}
printf("\n\nTotal << %d >> no of palindrom in given string\n",count);
return 0;
}

Let me know your thoughts.

To copy the contents of an array A to contents of Array B without using loops and any standard string copy functions.

2012-07-07T23:49:00.001+05:30

#include<stdio.h>
int Copy_Array(int A[],int B[], int n,int i)
{
if(i==n)
return 0;
B[i]=A[i];
Copy_Array(A,B,n,i+1);
return 0;
}

int main()
{
int A[10],B[10],i,n;
printf("Enter size : ");
scanf("%d",&n);
printf("Enter Array : ");
for(i=0;i<n;i++)
scanf("%d",&A[i]);
printf("\nArray A : ");
for(i=0;i<n;i++)
printf("%d ",A[i]);
Copy_Array(A,B,n,0);
printf("\nArray B : ");
for(i=0;i<n;i++)
printf("%d ",B[i]);
return 0;
}

Let me know your thoughts.

Verify if the given password is valid or not

2012-07-07T19:02:00.003+05:30

Verify if the given password is valid or not.
for a valid password it should follow these 2 condition
1. must be 5-12 characters long
2. must contain atleast one number and one lowercase & one upper character and a symbol

#include<stdio.h>
#include<string.h>
int main()
{
char S[20],c;
int i=0,l,L=0,U=0,N=0,Sy=0;
gets(S);
l=strlen(S);
if(l>12 || l<5)
{
printf("\n\Invalid password \n");
return 0;
}
while(S[i++]!=NULL)
if(S[i-1]>='a'&&S[i-1]<='z')
L=1;
else if(S[i-1]>='A'&&S[i-1]<='Z')
U=1;
else if(S[i-1]>='0'&&S[i-1]<='9')
N=1;
else Sy=1;
if(L&&U&&N&&Sy)
printf("\nvalid \n");
else printf("\n\Invalid password \n");
return 0;
}

Let me know your thoughts.

Find the seed of a number.

2012-07-07T17:57:00.001+05:30

Eg : 1716 = 14314*3 =1716so 143 is the seed of 1716.
find all possible seed for a given number.

// Input || OUTPUT
// 1250 125
// 174720 1456

#include<stdio.h>
int main()
{
int n,i,j,m,f=0;
scanf("%d",&n);
printf("\n");
for(i=0;i<n;i++)
{
m=i;
j=i;
while(j)
{
m=m*(j%10);
j=j/10;
}
if(m==n)
{
printf("\n%d",i);
f=1;
}
}
if(!f)
printf("\nNo such number exit\n\n");
return 0;
}

Find duplicates in O(n) time and O(1) extra space

2012-07-06T22:38:00.000+05:30

Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times.Find these repeating numbers in O(n) and using only constant memory space.

For example, let n be 7 and array be {1, 2, 3, 1, 3, 0, 6}, the answer should be 1 & 3.

#include<stdio.h>
#include<math.h>
int main()
{
int n,A[15],i,flage,cycle=0,j,temp;
printf("\nEnter size of array : ");
scanf("%d",&n);
printf("\nEnter elements of array( %d space saperated integer) : \n",n );
for(i=0;i<n;i++)
scanf("%d",&A[i]);
flage=0;j=0;
printf("\n OUTPUT : ");

for(i=0;i<n;i++)
{
if( A[abs(A[i])] >= 0)
A[abs(A[i])]=-A[abs(A[i])];
else
printf(" %d",abs(A[i]) );
}
return 0;
}

Let me know your thoughts.

Kahan We Met

2012-07-06T20:46:00.001+05:30

Geet is travelling to meet her boyfriend in Mumbai.

Aditya is travelling to Mumbai for his business meeting.

Geet starts from Delhi while Aditya starts from Bangalore. Both of them are travelling on bus passing from town to town.

They meet at a nondescript city and end up travelling together to Mumbai. By the end of the journey they fall in love and settle down.

Enough of Bollywood stories, The problem at hand is to find Kahan they met?

For all interested programmers, Represent this as a programming problem(Data Structure) and find the location where Geet and Aditya would have met.

P.S. No need to actually find any city on the map where they met.

// ===========================================================================

// Input Formate :

// 12 13 14 21 22 (these are 5 city pincode(2 disit integer pincoede for each city they pass through))

// 31 32 33 42 20 21 22 (these are 7 city pincode )

// Output : 21 (pincode of the city where they will met)

// ===================================================================

#include<stdio.h>

int kahan_they_met(int A[20],int G[20],int g,int a)
{
int i_g=0,i_a=0;
if(g>a)
i_g+=g-a;
else i_a+=a-g;

while(i_g <= g)
{
if(A[i_a]==G[i_g])
return A[i_a];
i_g++;
i_a++;
}
return 0;
}

int convert_input(int K[],char string[200])
{
int j=0,i=0,temp=0;
while(string[i]!='\0')
{
temp=0;
while(string[i]!=' ' && string[i]!='\0')
temp=temp*10 + (string[i++]-'0') ;
if(string[i]==' ')
i++;
K[j++]=temp;
}

return j-1;
}

int main()
{
char string[200];
int g,a,i,G[20],A[20],met;
printf("\nEnter city beetween D and M for GEET (eg: 12 13 14 21 22 ) : ");
gets(string);
g=convert_input(G,string);

printf("\n-----------------\nEnter city beetween B and M for for Aditya visit : ");
gets(string);
a=convert_input(A,string);

// for(i=0;i<=g;i++)
// printf("\n%d>>%d",i,G[i]);

met=kahan_they_met(A,G,g,a);

if(met)
printf("\n\nGeet and Aditya will met at : %d\n",met);
else
printf("\n\nGeet and Aditya will met at Mumbai\n");

return 0;

}

Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?

2012-07-06T15:28:00.001+05:30

INPUT > OUTPUT INPUT > > OUTPUT
4 4
1 2 3 4 1 1 1 1 11 12 13 14 41 31 21 11
1 2 3 4 2 2 2 2 21 22 23 24 42 32 22 12
1 2 3 4 3 3 3 3 31 32 33 34 43 33 23 13
1 2 3 4 4 4 4 4 41 42 43 44 44 34 24 14

#include<stdio.h>
void rotate(int I[10][10],int O[10][10],int j,int i)
{
int k,l,temp;
for(k=0;k<j;k++)
{
temp=I[k+i][i]; // copy of Left in temp
O[k+i][i]=I[i+j-1][i+k]; // copy bottom to left
O[i+j-1][i+k]=I[i+j-k-1][i+j-1]; // copy right to bottom
O[i+j-k-1][i+j-1]=I[i][i+j-k-1]; // copy top to right
O[i][i+j-k-1]=temp; // copy temp(left) to top
}
}// end of rotate

int main()
{
int I[10][10],O[10][10],i,j,N;
scanf("%d",&N);
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
scanf("%d",&I[i][j]);
}
i=0;
j=N;
while(j>0)
{
rotate(I,O,j,i); // i is base, j is size
j=j-2;
i++;
}
printf("\n\n");
for(i=0;i<N;i++)
{
printf("\n");
for(j=0;j<N;j++)
printf("%d ",O[i][j]);
}
return 0;
}

Reduce aabcccccaaa to a2b1c5a3 for new string is of smaller size otherwise live it original

2012-07-06T15:23:00.003+05:30

#include<stdio.h>
#include<string.h>
int main()
{
char s[20],o[20],temp;
int i=0,j=0,c=0;
gets(s);
while(s[i]!=NULL)
{
temp=s[i];
while(s[i]== temp)
{
i++;
c++;
}
o[j++]=temp;
o[j++]=c+48;// ASCII of 0 is 48
c=0;
}// end of while
if(strlen(o)<strlen(s))
puts(o);
else
puts(s);
return 0;
}

Write a method to decide if two strings are premutation of each other or not.

2012-07-06T15:02:00.000+05:30

#include<stdio.h>
#include<string.h>
int check_permutation(char s1[], char s2[])
{
int c[27],i;
for(i=0;i<27;i++)
c[i]=0;
if(strlen(s1)!=strlen(s2))
return 0;
for(i=0;s1[i]!=NULL;i++)
{
c[0]= s1[i];
if(c[0]>96)
c[0]=c[0]-96;
else c[0]=c[0]-64;
c[c[0]]++;

c[0]= s2[i];
if(c[0]>96)
c[0]=c[0]-96;
else c[0]=c[0]-64;
c[c[0]]--;
}
for(i=1;i<27;i++)
{
if(c[i]!=0)
return 0;
}
return 1;
}

int main()
{
char s1[20],s2[20];
scanf("%s%s",&s1,&s2);
if( check_permutation(s1,s2) )
printf("\n Permutation of each other");
else
printf("\n NOT a Permutation ");
return 0;
}

Implement an algorithm to determine if a string has all unique characters.

2012-07-06T14:56:00.000+05:30

#include<stdio.h>
#include<string.h>
int main()
{
char input[27];
int i,l,check[28];
scanf("%s",&input);
l=strlen(input);
for(i=0;i<27;i++)
check[i]=0;
for(i=0;i<l;i++)
{
check[27]=input[i];
if(check[27]>96)
check[27]=check[27]-96;
else
check[27]=check[27]-64;
if(!check[check[27]])
check[check[27]]=1;
else
{
printf("\nNOT");
return 0;
}
}
printf("\n unique");
return 0;
}

Input	Output
`3 1 4 10 120 100 1000`	`Case #1: 2 Case #2: 0 Case #3: 2`

Input	Output
`3 3 3 2 1 2 1 1 1 2 1 2 5 5 2 2 2 2 2 2 1 1 1 2 2 1 2 1 2 2 1 1 1 2 2 2 2 2 2 1 3 1 2 1`	`Case #1: YES Case #2: NO Case #3: YES`

CODES

PERL SCRIPT for killing CLOSE_WAIT

Install PhantomJS on Ubuntu

Some learning on Cassandra - I

Python script for get the files_from_directory, using_grep_on_file, on_day_increment

Google Code Jam Problem C. Fair and Square

Below is my solution for Problem C. Fair and Square

Problem

Solving this problem

Input

Output

Limits

Small dataset

First large dataset

Second large dataset

Sample

Google Code Jam Problem B. Lawnmower

Below is my solution for Problem B: Lawnmower

Problem

Problem

Input

Output

Limits

Small dataset

Large dataset

Sample

Google Code Jam Problem A. Tic-Tac-Toe-Tomek

Below is my solution for Problem A: Tic-Tac-Toe-Tomek

Problem

Input

Output

Limits

Small dataset

Large dataset

Sample

Note

write a function to check given string matches with given pattern

Finding Pythagorean triplets in an array.

Re-arrange the odd/even to odd/even places

Sort an Array of 0 1& 2 in one pass

Given a 2–d matrix , which has only 1’s and 0’s in it. Find the total number of connected sets in that matrix.

REVERSE a string and if any charecter is repeating then keep that one comes last

WAP to find the position of a given word in a given text

Which type of triangle is ?

Find the number of substrings of a string that are palindromes

To copy the contents of an array A to contents of Array B without using loops and any standard string copy functions.

Verify if the given password is valid or not

Find the seed of a number.

Eg : 1716 = 143*1*4*3 =1716so 143 is the seed of 1716.find all possible seed for a given number.

Find duplicates in O(n) time and O(1) extra space

Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times.Find these repeating numbers in O(n) and using only constant memory space.

For example, let n be 7 and array be {1, 2, 3, 1, 3, 0, 6}, the answer should be 1 & 3.

Kahan We Met

Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?

Reduce aabcccccaaa to a2b1c5a3 for new string is of smaller size otherwise live it original

Write a method to decide if two strings are premutation of each other or not.

Implement an algorithm to determine if a string has all unique characters.

Eg : 1716 = 14314*3 =1716so 143 is the seed of 1716.
find all possible seed for a given number.