Permutation of String letters: How to remove repeated permutations?

Question

Here is a standard function to print the permutations of characters of a 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++) //check till end of string
       {
          swap((a+i), (a+j));
          permute(a, i+1, n);
          swap((a+i), (a+j)); //backtrack
       }
   }
} 

void swap (char *x, char *y)
{
    char temp;
    temp = *x;
    *x = *y;
    *y = temp;
}

It works fine but there is a problem, it also prints some duplicate permutations, exapmle:

if string is "AAB"

the output is:

AAB
ABA
AAB
ABA
BAA
BAA

This is having 3 duplicate entries as well.

Can there be a way to prevent this to happen?

--

Thanks

Alok Kr.

Answer 1

Take notes which chars you swapped previously:

 char was[256];
 /*
 for(j = 0; j <= 255; j++)
    was[j] = 0;
 */
 bzero(was, 256);
 for (j = i; j <= n; j++)
 {
    if (!was[*(a+j)]) {
      swap((a+i), (a+j));
      permute(a, i+1, n);
      swap((a+i), (a+j)); //backtrack
      was[*(a+j)] = 1;
    }
 }

This has to be the fastest one from the entries so far, some benchmark on a "AAAABBBCCD" (100 loops):

native C             - real    0m0.547s
STL next_permutation - real    0m2.141s

Answer 2

Standard library has what you need:

#include <algorithm>
#include <iostream>
#include <ostream>
#include <string>
using namespace std;

void print_all_permutations(const string& s)
{
    string s1 = s;
    sort(s1.begin(), s1.end()); 
    do {
        cout << s1 << endl;
    } while (next_permutation(s1.begin(), s1.end()));
}

int main()
{
    print_all_permutations("AAB");
}

Result:

$ ./a.out
AAB
ABA
BAA