Home Practice Programming Print all permutations of a string with unique characters

Print all permutations of a string with unique characters

A permutation is an act of rearranging or reordering elements of a set or string etc.
For n elements, n! (n factorial) permutations are possible.
Ex-> Possible permutations of abc are abc, acb, bac, bca, cab, cba.

Here, given a string with n elements, we have to generate all possible permutation of this string.


  1. Start with the original string str and call the function find_permuation() with parameters original string, start index(0) and end index(str.size()-1).
  2. Iterate through every element of the string and perform the following operation i.e, for(i in range start to end-1).
  3. For every ith element of the string swap it with the 0th element, i.e, swap(str[start],str[i])
  4. Fix the current element, and again call the find_permuatation() function with parameters: the updated string, start index = current start index + 1 and original end index i.e, find_permuatation(str, start+1, end);
  5. After the current recursion terminates, re-swap the ith index with the 0th index i.e., swap(str[i],str[start])
  6. If start ==  end, we have reached the end of the recursion and found the permutation, print it and return
  7. Repeat step 2 to step 6 for every string element.

find_permutation(str, start, end){
    if(start == end)
    for i in range [start to end){
        swap(str[start], str[i]);
        //we fixed the ith index and now in next recursion
        //we will work with the remaining string elements
        find_permutation(str, start+1, end); 
        swap(str[i], str[start]);
   return 0;

Implementation of the above Algorithm in CPP

#include <bits/stdc++.h>
using namespace std;

int find_permutation(string str, int start, int end){


	for(int i=start; i<end; i++){
		find_permutation(str, start+1,end);
	return 0;

int main()
	string str = "abc";
	find_permutation(str, 0, str.size());
	return 0;



Time Complexity

The time complexity of the above algorithm is O(N*N!) where N is the size of the string.


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