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

Given an array of integers [3,1,4,5,7,6,10,8],
re arrange elements such that either even numbers in even locations or odd numbers in odd locations.
in O(n)

output :  6 1 4 5 8 3 10 7

#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;
}

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Sort an Array of 0 1& 2 in one pass

 Given an array int a[] = {0, 1, 1 ,0, 2, 2, 1, 0 ,1, 2, 1,0 }
to be converted into {0, 0, 0, 0, 1, 1, 1, 1, 1 ,2 ,2 ,2}
inplace and single pass

Input :  int A[] = {2,2,2,1,1,1,0,0,0,1,2,0};
Output:  0 0 0 0 1 1 1 1 2 2 2 2

#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;
}

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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.

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 :

4

4

0 0 1 0

1 0 1 0

0 1 0 0

1 1 1 1

4

1 0 0 1

0 0 0 0

0 1 1 0

1 0 0 1

5

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

8

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 :

1

3

3

9

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;

}

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

    INPUT           OUTPUT
    ABC             CBA
    ABBC            CBA
    NHHGFDGS        SGDFHN
    ABBCC           CBA

#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

WAP to find the position of a given word in a given text.
  INPUT  :
        Text : my name is kumar name
        Word : name
 OUTPUT :
         Position  : 3 17

#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;
}

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Which type of triangle is ?

Write a function that receives three integer inputs for the lengths of the sides of
a triangle and returns one of four values to determine the triangle type
(1=scalene, 2=isosceles, 3=equilateral, 4=error).

#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;
}

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