Distinct permutations of the string | Set 2

Print all distinct permutations of a string having duplicates.

Examples:

Input : ABCA
Output : AABC AACB ABAC ABCA ACBA 
         ACAB BAAC BACA BCAA CABA 
         CAAB CBAA

An algorithm to print all distinct permutations has already been discussed here. Here we’ll discuss one more approach to do the same. Recall first how we print permutations without any duplicates in the input string. It is given here. Let’s now take the case of the string “ABAC”. While generating permutations, let’s say we are at index = 0, and swap it with all elements after it. When we reach i=2, we see that in the string s[index…i-1], there was an index that is equal to s[i]. Thus, swapping it will produce repeated permutations. Thus, we don’t swap it. The below explains it better.

Illustration: Let us understand with the below example. 

i = 0 1 2 3
    A B A C
index = 0, s[0] = A
Start swapping s[index] with s[i] following it:
i = index + 1 = 1 

Since s[index] != s[i], swap and recur.

i = 2, s[index] == s[i], don't swap

i = 3,  s[index] != s[i], swap and recur. 

The Below code does the same. 

C++

// C++ program to distinct permutations of the string
#include <bits/stdc++.h>
using namespace std;

// Returns true if str[curr] does not matches with any of the
// characters after str[start]
bool shouldSwap(char str[], int start, int curr)
{
    for (int i = start; i < curr; i++)
        if (str[i] == str[curr])
            return 0;
    return 1;
}

// Prints all distinct permutations in str[0..n-1]
void findPermutations(char str[], int index, int n)
{
    if (index >= n) {
        cout << str << endl;
        return;
    }

    for (int i = index; i < n; i++) {

        // Proceed further for str[i] only if it
        // doesn't match with any of the characters
        // after str[index]
        bool check = shouldSwap(str, index, i);
        if (check) {
            swap(str[index], str[i]);
            findPermutations(str, index + 1, n);
            swap(str[index], str[i]);
        }
    }
}

// Driver code
int main()
{
    char str[] = "ABCA";
    int n = strlen(str);
    findPermutations(str, 0, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to distinct permutations of the string
#include <stdbool.h>
#include <stdio.h>
#include <string.h>

// Returns true if str[curr] does not matches with any of
// the characters after str[start]
bool shouldSwap(char str[], int start, int curr)
{
    for (int i = start; i < curr; i++)
        if (str[i] == str[curr])
            return 0;
    return 1;
}

// Prints all distinct permutations in str[0..n-1]
void findPermutations(char str[], int index, int n)
{
    if (index >= n) {
        printf("%s\n", str);
        return;
    }

    for (int i = index; i < n; i++) {

        // Proceed further for str[i] only if it
        // doesn't match with any of the characters
        // after str[index]
        bool check = shouldSwap(str, index, i);
        if (check) {
            // Swapping the str[index] with str[i]
            char temp = str[index];
            str[index] = str[i];
            str[i] = temp;
            findPermutations(str, index + 1, n);
            // Swapping the str[index] with str[i]
            temp = str[index];
            str[index] = str[i];
            str[i] = temp;
        }
    }
}

// Driver code
int main()
{
    char str[] = "ABCA";
    int n = strlen(str);
    findPermutations(str, 0, n);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to distinct permutations of the string
public class GFG {

// Returns true if str[curr] does not matches with any of the
// characters after str[start]
    static boolean shouldSwap(char str[], int start, int curr) {
        for (int i = start; i < curr; i++) {
            if (str[i] == str[curr]) {
                return false;
            }
        }
        return true;
    }

// Prints all distinct permutations in str[0..n-1]
    static void findPermutations(char str[], int index, int n) {
        if (index >= n) {
            System.out.println(str);
            return;
        }

        for (int i = index; i < n; i++) {

            // Proceed further for str[i] only if it
            // doesn't match with any of the characters
            // after str[index]
            boolean check = shouldSwap(str, index, i);
            if (check) {
                swap(str, index, i);
                findPermutations(str, index + 1, n);
                swap(str, index, i);
            }
        }
    }

    static void swap(char[] str, int i, int j) {
        char c = str[i];
        str[i] = str[j];
        str[j] = c;
    }

    // Driver code
    public static void main(String[] args) {

        char str[] = {'A', 'B', 'C', 'A'};
        int n = str.length;
        findPermutations(str, 0, n);
    }

}

Python3

# Python3 program to distinct
# permutations of the string

# Returns true if str[curr] does not
# matches with any of the characters
# after str[start]
def shouldSwap(string, start, curr):

    for i in range(start, curr):
        if string[i] == string[curr]:
            return 0
    return 1

# Prints all distinct permutations
# in str[0..n-1]
def findPermutations(string, index, n):

    if index >= n:
        print(''.join(string))
        return

    for i in range(index, n):

        # Proceed further for str[i] only
        # if it doesn't match with any of
        # the characters after str[index]
        check = shouldSwap(string, index, i)
        if check:
            string[index], string[i] = string[i], string[index]
            findPermutations(string, index + 1, n)
            string[index], string[i] = string[i], string[index]

# Driver code
if __name__ == "__main__":

    string = list("ABCA")
    n = len(string)
    findPermutations(string, 0, n)

# This code is contributed by Rituraj Jain

C#

// C# program to distinct permutations
// of the string
using System;

class GFG
{

// Returns true if str[curr] does
// not matches with any of the
// characters after str[start]
static bool shouldSwap(char []str,
                       int start, int curr)
{
    for (int i = start; i < curr; i++)
    {
        if (str[i] == str[curr])
        {
            return false;
        }
    }
    return true;
}

// Prints all distinct permutations
// in str[0..n-1]
static void findPermutations(char []str,
                             int index, int n)
{
    if (index >= n)
    {
        Console.WriteLine(str);
        return;
    }

    for (int i = index; i < n; i++)
    {

        // Proceed further for str[i] only
        // if it doesn't match with any of
        // the characters after str[index]
        bool check = shouldSwap(str, index, i);
        if (check)
        {
            swap(str, index, i);
            findPermutations(str, index + 1, n);
            swap(str, index, i);
        }
    }
}

static void swap(char[] str, int i, int j)
{
    char c = str[i];
    str[i] = str[j];
    str[j] = c;
}

// Driver code
public static void Main()
{
    char []str = {'A', 'B', 'C', 'A'};
    int n = str.Length;
    findPermutations(str, 0, n);
}
}

// This code is contributed
// by 29AjayKumar

Javascript

<script>

// Javascript program to distinct permutations of the string

// Returns true if str[curr] does not matches with any of the
// characters after str[start]
    function shouldSwap(str, start, curr) {
        for (let i = start; i < curr; i++) {
            if (str[i] == str[curr]) {
                return false;
            }
        }
        return true;
    }

// Prints all distinct permutations in str[0..n-1]
    function findPermutations(str, index, n) {
        if (index >= n) {
            document.write(str);
            document.write("<br/>")
            return;
        }


        for (let i = index; i < n; i++) {

            // Proceed further for str[i] only if it
            // doesn't match with any of the characters
            // after str[index]
            let check = shouldSwap(str, index, i);
            if (check) {
                swap(str, index, i);
                findPermutations(str, index + 1, n);
                swap(str, index, i);
            }
        }
    }

    function swap(str, i, j) {
        let c = str[i];
        str[i] = str[j];
        str[j] = c;
    }

// Driver code

        let str = ['A', 'B', 'C', 'A'];
        let n = str.length;
        findPermutations(str, 0, n);

</script>

Output

ABCA
ABAC
ACBA
ACAB
AACB
AABC
BACA
BAAC
BCAA
CBAA
CABA
CAAB

Better Approach:

Generate all distinct strings simply using some if conditions. The technique above uses an extra loop inside the recursion which causes a major time complexity cost. Instead, we can improve it by little pre-processing. We first sort the given string and then apply the below code.

Below is the implementation of the above idea:

C++

// C++ program to print all the permutation
// of the given string.
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;

// count of total permutations
int total = 0;

void permute(int i, string &s)
{
    // base case
    if (i == (s.length() - 1)) {
        cout << s << endl;
        total++;
        return;
    }


    // loop from j = 1 to length of string
    for (int j = i; j < s.length(); j++) {
        if (j>i && s[i] == s[j])
            continue;
        if (j>i && s[j-1] == s[j]) {
            continue;
        }

        // swap the elements
        swap(s[i], s[j]);

        // recursion call
        permute(i + 1, s);

          //backtrack
          swap(s[i], s[j]);
    }
}

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

    // sort
    sort(s.begin(), s.end());

    // Function call
    permute(0, s);
    cout << "Total distinct permutations = " << total
        << endl;
    return 0;
}

// This code is contributed by risingStark.

Java

// Java program to print all the permutation
// of the given String.
//include <algorithm>

//include <String>
import java.util.*;

class GFG{

// Count of total permutations
static int total = 0;

static void permute(int i, String s)
{

    // Base case
    if (i == (s.length() - 1))
    {
        System.out.print(s + "\n");
        total++;
        return;
    }

    char prev = '*';

    // Loop from j = 1 to length of String
    for(int j = i; j < s.length(); j++)
    {
        char []temp = s.toCharArray();
        if (j > i && temp[i] == temp[j])
            continue;
        if (prev != '*' && prev == s.charAt(j))
        {
            continue;
        }

        // Swap the elements
        temp = swap(temp,i,j);
        prev = s.charAt(j);

        // Recursion call
        permute(i + 1, String.valueOf(temp));
    }
}

static char[] swap(char []arr, int i, int j)
{
    char temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

static String sortString(String inputString)
{

    // Convert input string to char array
    char tempArray[] = inputString.toCharArray();

    // Sort tempArray
    Arrays.sort(tempArray);

    // Return new sorted string
    return new String(tempArray);
}

// Driver code
public static void main(String[] args)
{
    String s = "abca";

    // Sort
    s = sortString(s);

    // Function call
    permute(0, s);
    System.out.print("Total distinct permutations = " +
                     total +"\n");
}
}

// This code is contributed by Amit Katiyar

Python3

# Python program to print all the permutation
# of the given String.

# Count of total permutations
total = 0

def swap(arr, i, j):
    temp = arr[i]
    arr[i] = arr[j]
    arr[j] = temp
    return arr

def permute(i, s):
    global total

    # Base case
    if(i == (len(s) - 1)):
        print(s)
        total += 1
        return

    prev = '*'

    # Loop from j = 1 to length of String
    for j in range(i,len(s)):
        temp = list(s)
        if (j > i and temp[i] == temp[j]):
            continue
        if (prev != '*' and prev == s[j]):
            continue

        # Swap the elements
        temp = swap(temp,i,j)
        prev = s[j]

        # Recursion call
        permute(i + 1, "".join(temp))

def sortString(inputString):

    # Convert input string to char array
    tempArray = list(inputString)

    # Sort tempArray
    tempArray.sort()

    # Return new sorted string
    return "".join(tempArray)

# Driver code
s = "abca"

# Sort
s = sortString(s)

# Function call
permute(0, s)
print(f"Total distinct permutations = {total}")

# This code is contributed by shinjanpatra

C#

// C# program to print all the permutation
// of the given String.
using System;
using System.Collections.Generic;

class GFG {

    // Count of total permutations
    static int total = 0;

    static void permute(int i, string s)
    {

        // Base case
        if (i == (s.Length - 1)) {

            Console.Write(Convert.ToString(s) + "\n");
            total++;
            return;
        }

        char prev = '*';

        // Loop from j = 1 to length of String
        for (int j = i; j < s.Length; j++) {
            char[] temp = s.ToCharArray();
            if (j > i && temp[i] == temp[j])
                continue;
            if (prev != '*' && prev == s[j]) {
                continue;
            }

            // Swap the elements
            temp = swap(temp, i, j);
            prev = s[j];

            // Recursion call
            permute(i + 1, new string(temp));
        }
    }

    static char[] swap(char[] arr, int i, int j)
    {
        char temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }

    static string sortString(string inputString)
    {

        // Convert input string to char array
        char[] tempArray = inputString.ToCharArray();

        // Sort tempArray
        Array.Sort(tempArray);

        // Return new sorted string
        return new string(tempArray);
    }

    // Driver code
    public static void Main(string[] args)
    {
        string s = "abca";

        // Sort
        s = sortString(s);

        // Function call
        permute(0, s);
        Console.Write("Total distinct permutations = "
                      + total + "\n");
    }
}

// This code is contributed by phasing17

Javascript

<script>

// JavaScript program to print all the permutation
// of the given String.


// Count of total permutations
let total = 0;
function permute(i,s)
{
    // Base case
    if (i == (s.length - 1))
    {
        document.write(s + "<br>");
        total++;
        return;
    }

    let prev = '*';

    // Loop from j = 1 to length of String
    for(let j = i; j < s.length; j++)
    {
        let temp = s.split("");
        if (j > i && temp[i] == temp[j])
            continue;
        if (prev != '*' && prev == s[j])
        {
            continue;
        }

        // Swap the elements
        temp = swap(temp,i,j);
        prev = s[j];

        // Recursion call
        permute(i + 1, (temp).join(""));
    }
}

function swap(arr,i,j)
{
    let temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

function sortString(inputString)
{
    // Convert input string to char array
    let tempArray = inputString.split("");

    // Sort tempArray
    (tempArray).sort();

    // Return new sorted string
    return (tempArray).join("");
}

// Driver code
let s = "abca";

// Sort
s = sortString(s);

// Function call
permute(0, s);
document.write("Total distinct permutations = " +
                 total +"<br>");


// This code is contributed by rag2127

</script>

Output

aabc
aacb
abac
abca
acba
acab
baac
baca
bcaa
caba
caab
cbaa
Total distinct permutations = 12

Time complexity:  O(N*N!)