C++ Program To Print All Permutations Of A Given String

A permutation also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation. 

Source: Mathword(http://mathworld.wolfram.com/Permutation.html)

Below are the permutations of string ABC. 
ABC ACB BAC BCA CBA CAB

Here is a solution that is used as a basis in backtracking.

C++

// C++ program to print all permutations
// with duplicates allowed
#include <bits/stdc++.h>
using namespace std;

// 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(string a, int l, int r)
{
    // Base case
    if (l == r)
        cout<<a<<endl;
    else
    {
        // Permutations made
        for (int i = l; i <= r; i++)
        {
            // Swapping done
            swap(a[l], a[i]);

            // Recursion called
            permute(a, l+1, r);

            //backtrack
            swap(a[l], a[i]);
        }
    }
}

// Driver Code
int main()
{
    string str = "ABC";
    int n = str.size();
    permute(str, 0, n-1);
    return 0;
}
// This is code is contributed by rathbhupendra

Output:

ABC
ACB
BAC
BCA
CBA
CAB

Algorithm Paradigm: Backtracking 

Time Complexity: O(n*n!) Note that there are n! permutations and it requires O(n) time to print a permutation.

Auxiliary Space: O(r – l)

Note: The above solution prints duplicate permutations if there are repeating characters in the input string. Please see the below link for a solution that prints only distinct permutations even if there are duplicates in input.
Print all distinct permutations of a given string with duplicates.
Permutations of a given string using STL

Another approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
#include <string>
using namespace std;

void permute(string s,
             string answer)
{
    if(s.length() == 0)
    {
        cout << answer << "  ";
        return;
    }
    for(int i = 0;
            i < s.length(); i++)
    {
        char ch = s[i];
        string left_substr = s.substr(0, i);
        string right_substr = s.substr(i + 1);
        string rest = left_substr + right_substr;
        permute(rest , answer+ch);
    }
}

// Driver code
int main()
{
    string s;
    string answer = "";

    cout << "Enter the string : ";
    cin >> s;

    cout <<
    "All possible strings are : ";
    permute(s, answer);
    return 0;
}

Output:

Enter the string : abc
All possible strings are : abc  acb  bac  bca  cab  cba

Time Complexity: O(n*n!) The time complexity is the same as the above approach, i.e. there are n! permutations and it requires O(n) time to print a permutation.

Auxiliary Space: O(|s|)

Please refer complete article on Write a program to print all permutations of a given string for more details!