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;
}

Let me know your thoughts.

Consider you have a grid of size m x n. There are stones placed randomly in some of the squares of this grid. Design a way to find out minimum rectangular area which covers all the stones in this grid.

Smallest Rectangle

Input :(place 1 as rock other wise 0)
5 4
0 0 0 0
0 1 0 0
0 0 0 1
1 0 0 0
0 0 0 0

Output:   (1,0), (1,3), (3,0), (3,3)
*/

#include<stdio.h>
#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
int main()
{
    int m,n,A[10][10],i,j,i1=10,i2=0,j1=10,j2=0;
    printf("enter m & n (m*n) :");
    scanf("%d %d",&m,&n);
    printf("entet array (place 1 as rock other wise 0) :\n");
    for(i=0;i<m;i++)
        for(j=0;j<n;j++)
            scanf("%d",&A[i][j]);
    for(i=0;i<m;i++)
        for(j=0;j<n;j++)
            if(A[i][j]==1)
            {
                i1=MIN(i1,i);
                i2=MAX(i2,i);
                j1=MIN(j1,j);
                j2=MAX(j2,j);
            }
    printf("\n(%d,%d), (%d,%d), (%d,%d), (%d,%d)\n",i1,j1,i1,j2,i2,j1,i2,j2);
return 0;
}

The character 'a' to 'z' are encoded as 1 - 26. Given a string of digits, compute the number of valid decodings of the string. For example, both 'aa' and 'k' can be encoded as '11'. Hence num_valid_encodings('11') = 2.

      recursive 

#include <stdio.h>
#include <string.h>
int count_decode(char S[10],int n)
{
    int x1=S[n-1]-'0',x2=S[n-2]-'0',z;
    if(!n)
        return 0;
    else if(n==1)
    {
        return 1;
    }
    else
    {
        z=x2*10+x1;
      //  printf("z=%d ",z);
        if(z>26)
            return (count_decode(S,n-1));
        else if(n==2)
            return (count_decode(S,n-1) + count_decode(S,n-2)+1);
        else
            return (count_decode(S,n-1) + count_decode(S,n-2));
    }
}
int main()
{
    int i,j,n;
    char S[10];
    gets(S);
    n=strlen(S);
    printf("< %d >",count_decode(S,n));
return 0;
}

atoi is a function in the C programming language that converts a string into an integer numerical representation. atoi stands for ASCII to integer. It is included in the C standard library header file stdlib.h. Its prototype is as follows:

atoi : ASCII to integer 


    //INPUT : 24 83 0 2 


    //OUTPUT : 24 83 0 2

#include<stdio.h>

#include<string.h>
int atoi(int K[10],char string[30])
{
    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;
}

int main()
{
    char string[30];
    int i,a,A[10];
    printf("\nEnter String : ");
    gets(string);
    a=atoi(A,string);
    printf("\nIntegers are : ");
    for(i=0;i<a;i++)
        printf("%d ",A[i]);
    printf("\n");
return 0;
}

String Masking : Given a string"S" and we have to remove a an occurance of a small string "M" from there.

Input:
KUMAR
UM

Output:
KAR



#include<stdio.h>
#include<string.h>
int str_com(char K[20],int i,int m,char M[10])
{
    int k=0;
    for(k=0;k<m;k++)
        if(K[i+k]!=M[k])
            break;
    if(k==m)
        return 1;
    return 0;
}

int main()
{
    char S[20],M[10];
    int i,j=0,l,m,k,flage;
    gets(S);
    gets(M);
    l=strlen(S);
    m=strlen(M);
    for(i=0;i<l;i++)
    {
        if(S[i]!=M[0])
            S[j++]=S[i];
        else
            if( str_com(S,i,m,M) )
                i+=m-1;
            else S[j++]=S[i];
    }
    S[j]=NULL;
    printf("< %s >",S);
return 0;
}