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;
}
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));
}
}
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));
}
}
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));
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 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.
# 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
<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>
Time Complexity: O(n)
Auxiliary Space: O(n)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem