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.

sort an array a[n] containing 0s and 1s with time complexity n and you cant use more than 1 data structure

//      TIME: O(n)  ||   Space : O(1)


#include<stdio.h>
int main()
{
    int A[20],i,j,n,temp;
    printf("Enter n :");
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%d",&A[i]);
    printf("\n--");
    i=0;
    j=n-1;
    while(i<j)
    {
        while(A[i]==0)
            i++;
        while(A[j]==1)
            j--;
        if(i<j)
        {
            temp=A[i];
            A[i]=A[j];
            A[j]=temp;
            i++; j--;
        }
    }// end of while
    printf("\n");
    for(i=0;i<n;i++)
        printf("%d ",A[i]);
    printf("\n");
return 0;
}