# Print all distinct permutations of a given string with duplicates

Given a string that may contain duplicates, write a function to print all permutations of given string such that no permutation is repeated in output.

Examples:

Input: str[] = "AB" Output: AB BA Input: str[] = "AA" Output: AA Input: str[] = "ABC" Output: ABC ACB BAC BCA CBA CAB Input: str[] = "ABA" Output: ABA AAB BAA Input: str[] = "ABCA" Output: AABC AACB ABAC ABCA ACBA ACAB BAAC BACA BCAA CABA CAAB CBAA

We have discussed an algorithm to print all permutations in below post. It is strongly recommended to refer below post as a prerequisite of this post.

Write a C program to print all permutations of a given string

The algorithm discussed on above link doesn’t handle duplicates.

`// Program to print all permutations of a ` `// string in sorted order. ` `#include <stdio.h> ` `#include <stdlib.h> ` `#include <string.h> ` ` ` `/* Following function is needed for library ` ` ` `function qsort(). */` `int` `compare(` `const` `void` `* a, ` `const` `void` `* b) ` `{ ` ` ` `return` `(*(` `char` `*)a - *(` `char` `*)b); ` `} ` ` ` `// A utility function two swap two characters ` `// a and b ` `void` `swap(` `char` `* a, ` `char` `* b) ` `{ ` ` ` `char` `t = *a; ` ` ` `*a = *b; ` ` ` `*b = t; ` `} ` ` ` `// This function finds the index of the ` `// smallest character which is greater ` `// than 'first' and is present in str[l..h] ` `int` `findCeil(` `char` `str[], ` `char` `first, ` `int` `l, ` `int` `h) ` `{ ` ` ` `// initialize index of ceiling element ` ` ` `int` `ceilIndex = l; ` ` ` ` ` `// Now iterate through rest of the ` ` ` `// elements and find the smallest ` ` ` `// character greater than 'first' ` ` ` `for` `(` `int` `i = l + 1; i <= h; i++) ` ` ` `if` `(str[i] > first && str[i] < str[ceilIndex]) ` ` ` `ceilIndex = i; ` ` ` ` ` `return` `ceilIndex; ` `} ` ` ` `// Print all permutations of str in sorted order ` `void` `sortedPermutations(` `char` `str[]) ` `{ ` ` ` `// Get size of string ` ` ` `int` `size = ` `strlen` `(str); ` ` ` ` ` `// Sort the string in increasing order ` ` ` `qsort` `(str, size, ` `sizeof` `(str[0]), compare); ` ` ` ` ` `// Print permutations one by one ` ` ` `bool` `isFinished = ` `false` `; ` ` ` `while` `(!isFinished) { ` ` ` ` ` `// print this permutation ` ` ` `static` `int` `x = 1; ` ` ` `printf` `(` `"%d %s \n"` `, x++, str); ` ` ` ` ` `// Find the rightmost character ` ` ` `// which is smaller than its next ` ` ` `// character. Let us call it 'first ` ` ` `// char' ` ` ` `int` `i; ` ` ` `for` `(i = size - 2; i >= 0; --i) ` ` ` `if` `(str[i] < str[i + 1]) ` ` ` `break` `; ` ` ` ` ` `// If there is no such character, all ` ` ` `// are sorted in decreasing order, ` ` ` `// means we just printed the last ` ` ` `// permutation and we are done. ` ` ` `if` `(i == -1) ` ` ` `isFinished = ` `true` `; ` ` ` `else` `{ ` ` ` ` ` `// Find the ceil of 'first char' ` ` ` `// in right of first character. ` ` ` `// Ceil of a character is the ` ` ` `// smallest character greater ` ` ` `// than it ` ` ` `int` `ceilIndex = findCeil(str, ` ` ` `str[i], i + 1, size - 1); ` ` ` ` ` `// Swap first and second characters ` ` ` `swap(&str[i], &str[ceilIndex]); ` ` ` ` ` `// Sort the string on right of 'first char' ` ` ` `qsort` `(str + i + 1, size - i - 1, ` ` ` `sizeof` `(str[0]), compare); ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver program to test above function ` `int` `main() ` `{ ` ` ` `char` `str[] = ` `"ACBC"` `; ` ` ` `sortedPermutations(str); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

1 ABCC 2 ACBC 3 ACCB 4 BACC 5 BCAC 6 BCCA 7 CABC 8 CACB 9 CBAC 10 CBCA 11 CCAB 12 CCBA

The above code is taken from a comment below by Mr. Lazy.

Time Complexity: O(n^{2} * n!)

Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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