Count of distinct permutation of a String obtained by swapping only unequal characters

Last Updated : 23 Jul, 2025

Given a string find the number of unique permutations that can be obtained by swapping two indices such that the elements at these indices are distinct.

NOTE: Swapping is always performed in the original string.

Examples:

Input: str = "sstt"
Output: 5
Explanation:
Swap str[0] with str[2], string obtained "tsst" which is valid (str[0]!=str[2])
Swap str[0] with str[3]. string obtained "tsts"
Swap str[1] with str[2], string obtained "stst"
Swap str[1] with str[3], string obtained "stts"
Hence total 5 strings can be obtained including the given string ("sstt")

Input: str = "abcd"
Output: 7

Naive Approach:

Create a set to store unique strings
Loop through each pair of indices in the given string
Check if the elements at the pair of indices are distinct
If the elements are distinct, swap the elements at those indices and add the resulting string to the set
Swap the elements back to their original positions
Add 1 to count the original string
Return the number of unique permutations, including the original string.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <set>
#include <string>

using namespace std;

int countUniquePermutations(string str)
{
    // create a set to store unique strings
    set<string> uniqueStrings;

    // loop through each pair of indices
    // in the given string
    for (int i = 0; i < str.length(); i++) {
        for (int j = i + 1; j < str.length(); j++) {
            // check if the elements at the
            // pair of indices are distinct

            if (str[i] != str[j]) {
                swap(str[i], str[j]);
                uniqueStrings.insert(str);
                swap(str[i], str[j]);
            }
        }
    }
    // add 1 to count the original string
    // and return number of unique permutations
    return uniqueStrings.size() + 1;
}

int main()
{
    string str = "sstt";
    cout << countUniquePermutations(str);

    return 0;
}

Java

import java.util.HashSet;
import java.util.Set;

public class UniquePermutations {
    public static int countUniquePermutations(String str)
    {
        // Create a set to store unique strings
        Set<String> uniqueStrings = new HashSet<>();

        // Loop through each pair of indices in the given
        // string
        for (int i = 0; i < str.length(); i++) {
            for (int j = i + 1; j < str.length(); j++) {
                // Check if the elements at the pair of
                // indices are distinct
                if (str.charAt(i) != str.charAt(j)) {
                    // Swap the characters at the indices
                    char[] strArray = str.toCharArray();
                    char temp = strArray[i];
                    strArray[i] = strArray[j];
                    strArray[j] = temp;
                    String swappedStr
                        = new String(strArray);

                    // Add the swapped string to the set
                    uniqueStrings.add(swappedStr);

                    // Swap back to the original string
                    temp = strArray[i];
                    strArray[i] = strArray[j];
                    strArray[j] = temp;
                }
            }
        }

        // Add 1 to count the original string and return the
        // number of unique permutations
        return uniqueStrings.size() + 1;
    }

    public static void main(String[] args)
    {
        String str = "sstt";
        System.out.println(countUniquePermutations(str));
    }
}

Python3

def count_unique_permutations(s):
    # Create a set to store unique strings
    unique_strings = set()

    # Loop through each pair of indices in the given string
    for i in range(len(s)):
        for j in range(i + 1, len(s)):
            # Check if the elements at the pair of indices are distinct
            if s[i] != s[j]:
                # Swap the elements at the indices to generate a new string
                # and add it to the set
                s_list = list(s)
                s_list[i], s_list[j] = s_list[j], s_list[i]
                unique_strings.add("".join(s_list))

    # Add 1 to count the original string and return the number of unique permutations
    return len(unique_strings) + 1

# Driver Code
if __name__ == "__main__":
    input_str = "sstt"
    print(count_unique_permutations(input_str))

using System;
using System.Collections.Generic;

class Program
{
    static int CountUniquePermutations(string str)
    {
        // Create a HashSet to store unique strings
        HashSet<string> uniqueStrings = new HashSet<string>();

        // Loop through each pair of indices in the given string
        for (int i = 0; i < str.Length; i++)
        {
            for (int j = i + 1; j < str.Length; j++)
            {
                // Check if the elements at the pair of indices are distinct
                if (str[i] != str[j])
                {
                    char[] chars = str.ToCharArray();
                    // Swap the elements at indices i and j
                    char temp = chars[i];
                    chars[i] = chars[j];
                    chars[j] = temp;
                    string newString = new string(chars);
                    
                    // Insert the unique string into the HashSet
                    uniqueStrings.Add(newString);

                    // Swap back to restore the original string
                    temp = chars[i];
                    chars[i] = chars[j];
                    chars[j] = temp;
                }
            }
        }

        // Add 1 to count the original string and return the number of unique permutations
        return uniqueStrings.Count + 1;
    }

    static void Main()
    {
        string str = "sstt";
        Console.WriteLine(CountUniquePermutations(str));
    }
}

JavaScript

function countUniquePermutations(str) {
    // Create a Set to store unique strings
    let uniqueStrings = new Set();

    // Loop through each pair of indices in the given string
    for (let i = 0; i < str.length; i++) {
        for (let j = i + 1; j < str.length; j++) {
            // Check if the elements at the pair of indices are distinct
            if (str[i] !== str[j]) {
                // Swap the characters at indices i and j
                let strArray = str.split('');
                [strArray[i], strArray[j]] = [strArray[j], strArray[i]];
                let modifiedStr = strArray.join('');
                uniqueStrings.add(modifiedStr);

                // Swap the characters back to their original positions
                [strArray[i], strArray[j]] = [strArray[j], strArray[i]];
            }
        }
    }
    
    // Add 1 to count the original string and return the number of unique permutations
    return uniqueStrings.size + 1;
}

let str = "sstt";
console.log(countUniquePermutations(str));

Output

Time Complexity: The time complexity of this algorithm is O(n² * log n), where n is the length of the input string. This is because we have two nested loops, and within each loop, we perform a constant-time set insertion operation, which has a time complexity of O(log n) on average. Therefore, the overall time complexity is O(n^2 * log n).

Auxiliary Space: The space complexity of this algorithm is also O(n²). This is because we create a set to store the unique strings, and the size of the set can be as large as n^2 in the worst case, if all possible string permutations are unique. Additionally, we store the input string itself, which has a size of n. Therefore, the overall space complexity is O(n^2 + n), which simplifies to O(n^2).

Efficient Approach: The problem can be solved using HashMap by the following steps:

Create a hashmap and store the frequency of every character of the given string.
Create a variable count, to store the total number of characters in the given string, i.e. count=str.length() and a variable ans to store the number of unique permutations possible and initialize ans=0.
Traverse the string and for each character:
- Find the number of different characters present in the right of the current index. This can be done by subtracting the frequency of that character by the total count.
- Now add this calculated value to ans, as this is the number of characters with which the current character can be swapped to create a unique permutation.
- Now, Reduce the frequency of the current character and count by 1, so that it cannot interfere with the calculations of the same elements present to the right of it.
Return ans+1, because the given string is also a unique permutation.

C++

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

// Function to calculate total
// number of valid permutations
int validPermutations(string str)
{
    unordered_map<char, int> m;

    // Creating count which is equal to the
    // Total number of characters present and
    // ans that will store the number of unique
    // permutations
    int count = str.length(), ans = 0;

    // Storing frequency of each character
    // present in the string
    for (int i = 0; i < str.length(); i++) {
        m[str[i]]++;
    }
    for (int i = 0; i < str.length(); i++) {
        // Adding count of characters by excluding
        // characters equal to current char
        ans += count - m[str[i]];

        // Reduce the frequency of the current character
        // and count by 1, so that it cannot interfere
        // with the calculations of the same elements
        // present to the right of it.
        m[str[i]]--;
        count--;
    }

    // Return ans+1 (Because the given string
    // is also a unique permutation)
    return ans + 1;
}

// Driver Code
int main()
{
    string str = "sstt";
    cout << validPermutations(str);
    return 0;
}

Java

// Java program for the above approach

// Importing HashMap class
import java.util.HashMap;

class GFG {

    // Function to calculate total
    // number of valid permutations
    static int validPermutations(String str)
    {

        HashMap<Character, Integer> m
            = new HashMap<Character, Integer>();

        // Creating count which is equal to the
        // Total number of characters present and
        // ans that will store the number of unique
        // permutations
        int count = str.length(), ans = 0;

        // Storing frequency of each character
        // present in the string
        for (int i = 0; i < str.length(); i++) {
            m.put(str.charAt(i),
                  m.getOrDefault(str.charAt(i), 0) + 1);
        }

        for (int i = 0; i < str.length(); i++) {
            // Adding count of characters by excluding
            // characters equal to current char
            ans += count - m.get(str.charAt(i));

            // Reduce the frequency of the current character
            // and count by 1, so that it cannot interfere
            // with the calculations of the same elements
            // present to the right of it.
            m.put(str.charAt(i), m.get(str.charAt(i)) - 1);
            count--;
        }

        // Return ans+1 (Because the given string
        // is also a unique permutation)
        return ans + 1;
    }

    public static void main(String[] args)
    {
        String str = "sstt";

        System.out.println(validPermutations(str));
    }
}

// This code is contributed by rajsanghavi9.

Python3

# Python 3 program for the above approach

# Function to calculate total
# number of valid permutations
def validPermutations(str):
    m = {}

    # Creating count which is equal to the
    # Total number of characters present and
    # ans that will store the number of unique
    # permutations
    count = len(str)
    ans = 0

    # Storing frequency of each character
    # present in the string
    for i in range(len(str)):
        if(str[i] in m):
            m[str[i]] += 1
        else:
            m[str[i]] = 1
    for i in range(len(str)):
      
        # Adding count of characters by excluding
        # characters equal to current char
        ans += count - m[str[i]]

        # Reduce the frequency of the current character
        # and count by 1, so that it cannot interfere
        # with the calculations of the same elements
        # present to the right of it.
        m[str[i]] -= 1
        count -= 1

    # Return ans+1 (Because the given string
    # is also a unique permutation)
    return ans + 1

# Driver Code
if __name__ == '__main__':
    str = "sstt"
    print(validPermutations(str)) 

    # This code is contributed by SURENDRA_GANGWAR.

// C# program for the above approach

// Importing Dictionary class
using System;
using System.Collections.Generic;


public class GFG {

    // Function to calculate total
    // number of valid permutations
    static int validPermutations(String str)
    {

        Dictionary<char, int> m
            = new Dictionary<char, int>();

        // Creating count which is equal to the
        // Total number of characters present and
        // ans that will store the number of unique
        // permutations
        int count = str.Length, ans = 0;

        // Storing frequency of each character
        // present in the string
        for (int i = 0; i < str.Length; i++) {
            if(m.ContainsKey(str[i]))
                m[str[i]]=m[str[i]]+1;
            else
                m.Add(str[i], 1);
        }

        for (int i = 0; i < str.Length; i++) {
            // Adding count of characters by excluding
            // characters equal to current char
            ans += count - m[str[i]];

            // Reduce the frequency of the current character
            // and count by 1, so that it cannot interfere
            // with the calculations of the same elements
            // present to the right of it.
            if(m.ContainsKey(str[i]))
                m[str[i]]=m[str[i]]-1;
            count--;
        }

        // Return ans+1 (Because the given string
        // is also a unique permutation)
        return ans + 1;
    }

    public static void Main(String[] args)
    {
        String str = "sstt";

        Console.WriteLine(validPermutations(str));
    }
}

// This code is contributed by shikhasingrajput

JavaScript

<script>

// JavaScript program for the above approach

// Function to calculate total
// number of valid permutations
function validPermutations(str) {
  let m = new Map();

  // Creating count which is equal to the
  // Total number of characters present and
  // ans that will store the number of unique
  // permutations
  let count = str.length,
    ans = 0;

  // Storing frequency of each character
  // present in the string
  for (let i = 0; i < str.length; i++) {
    if (m.has(str[i])) {
      m.set(str[i], m.get(str[i]) + 1);
    } else {
      m.set(str[i], 1);
    }
  }
  for (let i = 0; i < str.length; i++)
  {
  
    // Adding count of characters by excluding
    // characters equal to current char
    ans += count - m.get(str[i]);


    // Reduce the frequency of the current character
    // and count by 1, so that it cannot interfere
    // with the calculations of the same elements
    // present to the right of it.
    m.set(str[i], m.get(str[i]) - 1);
    count--;
  }

  // Return ans+1 (Because the given string
  // is also a unique permutation)
  return ans + 1;
}

// Driver Code
let str = "sstt";
document.write(validPermutations(str));

</script>

Output

Time Complexity: O(n)
Auxiliary Space: O(n)

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