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Permutations of given String

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Given a string S, the task is to write a 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! permutations. 

Examples:

Input: S = “ABC”
Output: “ABC”, “ACB”, “BAC”, “BCA”, “CBA”, “CAB”

Input: S = “XY”
Output: “XY”, “YX”

Print permutations of a given string using backtracking:

Recursion Tree for permutations of string "ABC"

Recursion Tree for permutations of string “ABC”

Follow the given steps to solve the problem:

  • Create a function permute() with parameters as input string, starting index of the string, ending index of the string
  • Call this function with values input string, 0, size of string – 1
    • In this function, if the value of  L and R is the same then print the same string
      • Else run a for loop from L to R and swap the current element in the for loop with the inputString[L]
      • Then again call this same function by increasing the value of L by 1
      • After that again swap the previously swapped values to initiate backtracking

Below is the implementation of the above approach:

C++14




// 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();
  
    // Function call
    permute(str, 0, n - 1);
    return 0;
}
  
// This is code is contributed by rathbhupendra


C




// C program to print all permutations with duplicates
// allowed
#include <stdio.h>
#include <string.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 l, int r)
{
    int i;
    if (l == r)
        printf("%s\n", a);
    else {
        for (i = l; i <= r; i++) {
            swap((a + l), (a + i));
            permute(a, l + 1, r);
            swap((a + l), (a + i)); // backtrack
        }
    }
}
  
/* Driver code */
int main()
{
    char str[] = "ABC";
    int n = strlen(str);
    permute(str, 0, n - 1);
    return 0;
}


Java




// Java program to print all permutations of a
// given string.
public class Permutation {
  
    // Function call
    public static void main(String[] args)
    {
        String str = "ABC";
        int n = str.length();
        Permutation permutation = new Permutation();
        permutation.permute(str, 0, n - 1);
    }
  
    /**
     * permutation function
     * @param str string to calculate permutation for
     * @param l starting index
     * @param r end index
     */
    private void permute(String str, int l, int r)
    {
        if (l == r)
            System.out.println(str);
        else {
            for (int i = l; i <= r; i++) {
                str = swap(str, l, i);
                permute(str, l + 1, r);
                str = swap(str, l, i);
            }
        }
    }
  
    /**
     * Swap Characters at position
     * @param a string value
     * @param i position 1
     * @param j position 2
     * @return swapped string
     */
    public String swap(String a, int i, int j)
    {
        char temp;
        char[] charArray = a.toCharArray();
        temp = charArray[i];
        charArray[i] = charArray[j];
        charArray[j] = temp;
        return String.valueOf(charArray);
    }
}
  
// This code is contributed by Mihir Joshi


Python3




# Python3 program to print all permutations with
# duplicates allowed
  
  
def toString(List):
    return ''.join(List)
  
# 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.
  
  
def permute(a, l, r):
    if l == r:
        print(toString(a))
    else:
        for i in range(l, r):
            a[l], a[i] = a[i], a[l]
            permute(a, l+1, r)
            a[l], a[i] = a[i], a[l]  # backtrack
  
  
# Driver code
string = "ABC"
n = len(string)
a = list(string)
  
# Function call
permute(a, 0, n)
  
# This code is contributed by Bhavya Jain


C#




// C# program to print all
// permutations of a given string.
using System;
  
class GFG {
    /**
    * permutation function
    * @param str string to
    calculate permutation for
    * @param l starting index
    * @param r end index
    */
    private static void permute(String str, int l, int r)
    {
        if (l == r)
            Console.WriteLine(str);
        else {
            for (int i = l; i <= r; i++) {
                str = swap(str, l, i);
                permute(str, l + 1, r);
                str = swap(str, l, i);
            }
        }
    }
  
    /**
     * Swap Characters at position
     * @param a string value
     * @param i position 1
     * @param j position 2
     * @return swapped string
     */
    public static String swap(String a, int i, int j)
    {
        char temp;
        char[] charArray = a.ToCharArray();
        temp = charArray[i];
        charArray[i] = charArray[j];
        charArray[j] = temp;
        string s = new string(charArray);
        return s;
    }
  
    // Driver Code
    public static void Main()
    {
        String str = "ABC";
        int n = str.Length;
        permute(str, 0, n - 1);
    }
}
  
// This code is contributed by mits


Javascript




<script>
// Javascript program to print all permutations of a
// given string.
  
function permute(str, l, r)
{
    if (l == r)
            document.write(str+"<br>");
        else
        {
            for (let i = l; i <= r; i++)
            {
                str = swap(str, l, i);
                permute(str, l + 1, r);
                str = swap(str, l, i);
            }
        }
}
  
function swap(a, i, j)
{
    let temp;
let charArray = a.split("");
temp = charArray[i] ;
charArray[i] = charArray[j];
charArray[j] = temp;
return (charArray).join("");
}
  
let str = "ABC";
let n = str.length;
permute(str, 0, n-1);
  
// This code is contributed by avanitrachhadiya2155
</script>


PHP




<?php 
// PHP program to print all 
// permutations of a given string. 
  
  
/** 
* permutation function 
* @param str string to 
* calculate permutation for 
* @param l starting index 
* @param r end index 
*/
function permute($str, $l, $r
    if ($l == $r
        echo $str. "\n"
    else
    
        for ($i = $l; $i <= $r; $i++) 
        
            $str = swap($str, $l, $i); 
            permute($str, $l + 1, $r); 
            $str = swap($str, $l, $i); 
        
    
  
/** 
* Swap Characters at position 
* @param a string value 
* @param i position 1 
* @param j position 2 
* @return swapped string 
*/
function swap($a, $i, $j
    $temp
    $charArray = str_split($a); 
    $temp = $charArray[$i] ; 
    $charArray[$i] = $charArray[$j]; 
    $charArray[$j] = $temp
    return implode($charArray); 
  
// Driver Code 
$str = "ABC"
$n = strlen($str); 
permute($str, 0, $n - 1); 
  
// This code is contributed by mits. 
?>


Output

ABC
ACB
BAC
BCA
CBA
CAB

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.



Last Updated : 18 Oct, 2023
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