All possible permutations of a string when duplicates are present??

Given a string how can we print all permutations of a string when duplicates are present??

I know the solution when no duplicates are present using back tracking

# include <stdio.h>

/* Function to swap values at two pointers */
void swap (char *x, char *y)
{
    char temp;
    temp = *x;
    *x = *y;
    *y = temp;
}
 
/* Function to print permutations of string
   This function takes three parameters:
   1. String
   2. Starting index of the string
   3. Ending index of the string. */
void permute(char *a, int i, int n) 
{
   int j; 
   if (i == n)
     printf("%s\n", a);
   else
   {
        for (j = i; j <= n; j++)
       {
          swap((a+i), (a+j));
          permute(a, i+1, n);
          swap((a+i), (a+j)); //backtrack
       }
   }
} 

/* Driver program to test above functions */
int main()
{
   char a[] = "ABC";  
   permute(a, 0, 2);
   getchar();
   return 0;
}

but using the same algorithm i end up printing same string more than once when duplicates are present?? Is there any way to print all permutations of string when duplicates are present :slight_smile:

you can have a look at this SO link…:slight_smile:

this is another good link…!!!

2 Likes

Perfect :smiley: And Now U can close this question :slight_smile: If u know wat I mean :slight_smile:

closed it…:stuck_out_tongue:

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