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Write a program to print all permutations of a given string

  • Difficulty Level : Medium
  • Last Updated : 09 Oct, 2021

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)

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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.



NewPermutation

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
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 program to test above functions */
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
{
    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
Python



# Python 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 xrange(l,r+1):
            a[l], a[i] = a[i], a[l]
            permute(a, l+1, r)
            a[l], a[i] = a[i], a[l] # backtrack
 
# Driver program to test the above function
string = "ABC"
n = len(string)
a = list(string)
permute(a, 0, n-1)
 
# 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
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.
?>
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>
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++



#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);
    }
 
}
 
int main()
{
    string s;
    string answer="";
 
    cout<<"Enter the string : ";
    cin>>s;
 
    cout<<"\nAll possible strings are : ";
    permute(s , answer);
    return 0;
}
Java



import java.util.*;
 
class GFG{
  
static void permute(String s , String answer)
{  
    if (s.length() == 0)
    {
        System.out.print(answer + "  ");
        return;
    }
     
    for(int i = 0 ;i < s.length(); i++)
    {
        char ch = s.charAt(i);
        String left_substr = s.substring(0, i);
        String right_substr = s.substring(i + 1);
        String rest = left_substr + right_substr;
        permute(rest, answer + ch);
    }
}
 
// Driver code
public static void main(String args[])
{
    Scanner scan = new Scanner(System.in);
     
    String s;
    String answer="";
     
    System.out.print("Enter the string : ");
    s = scan.next();
     
    System.out.print("\nAll possible strings are : ");
    permute(s, answer);
}
}
 
// This code is contributed by adityapande88
Python3



def permute(s, answer):
    if (len(s) == 0):
        print(answer, end = "  ")
        return
     
    for i in range(len(s)):
        ch = s[i]
        left_substr = s[0:i]
        right_substr = s[i + 1:]
        rest = left_substr + right_substr
        permute(rest, answer + ch)
 
# Driver Code
answer = ""
 
s = input("Enter the string : ")
 
print("All possible strings are : ")
permute(s, answer)
 
# This code is contributed by Harshit Srivastava
C#



using System;
 
public class GFG{
  
static void permute(String s , String answer)
{  
    if (s.Length == 0)
    {
        Console.Write(answer + "  ");
        return;
    }
     
    for(int i = 0 ;i < s.Length; i++)
    {
        char ch = s[i];
        String left_substr = s.Substring(0, i);
        String right_substr = s.Substring(i + 1);
        String rest = left_substr + right_substr;
        permute(rest, answer + ch);
    }
}
 
// Driver code
public static void Main(String []args)
{
     
     
    String s;
    String answer="";
     
    Console.Write("Enter the string : ");
    s = Console.ReadLine();
     
    Console.Write("\nAll possible strings are : ");
    permute(s, answer);
}
}
 
// This code is contributed by gauravrajput1
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 write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
 
 



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