Print all Unique permutations of a given string.
Last Updated :
23 Jul, 2025
Given a string that may contain duplicates, the task is find all unique permutations of given string in any order.
Examples:
Input: "ABC"
Output: ["ABC", "ACB", "BAC", "BCA", "CAB", "CBA"]
Explanation: Given string ABC has 6 unique permutations as "ABC", "ACB", "BAC", "BCA", "CAB" and "CBA".
Input: "AAA"
Output: ["AAA"]
Explanation: No other unique permutations can be formed as all the characters are same.
[Naive Approach] Generating All Permutations
The idea is to generate all possible permutations of a given string and store them in a hash set. The generated permutations were stored in hash set, to ensures that only unique permutations are retained as hashset does not allow duplicates. Once all permutations have been generated, we transfer the permutations from hash set to the result array.
This method is particularly useful when dealing with input strings that contain duplicate characters. Instead of manually handling the duplicates during the permutation generation, the use of a unordered set simplifies the process by eliminating duplicates automatically.
CPP
// C++ program to print unique permutation of string using hash set.
#include <algorithm>
#include <iostream>
#include <string>
#include <unordered_set>
#include <vector>
using namespace std;
// Function genPermutation is used to generate all possible permutaion.
void genPermutation(int i, string &s, vector<bool> &used,
string &curr, unordered_set<string>& st) {
if (i == s.size()) {
// Add the permutation to the result set
st.insert(curr);
return;
}
for (int j = 0; j < s.size(); j++) {
if (!used[j]) {
// Mark the character as used
used[j] = true;
curr.push_back(s[j]);
// Recurse with the next character
genPermutation(i + 1, s, used, curr,st);
// Backtrack and unmark the character
used[j] = false;
curr.pop_back();
}
}
}
vector<string> findPermutation(string &s) {
// To track if a character is used
vector<bool> used(s.size(), false);
unordered_set<string> st;
string curr = "";
// Start the recursion
genPermutation(0, s, used, curr, st);
// Convert the set to a vector
vector<string> res(st.begin(), st.end());
return res;
}
int main() {
string s = "ABC";
vector<string> res = findPermutation(s);
// Print the permutations
for (string perm: res)
cout << perm << " ";
return 0;
}
// Java program to print unique permutation of string using hash set.
import java.util.ArrayList;
import java.util.HashSet;
class GfG {
// Function genPermutation is used to generate all possible permutation.
static void genPermutation(int i, String s, boolean[] used,
StringBuilder curr, HashSet<String> st) {
if (i == s.length()) {
// Add the permutation to the result set
st.add(curr.toString());
return;
}
for (int j = 0; j < s.length(); j++) {
if (!used[j]) {
// Mark the character as used
used[j] = true;
curr.append(s.charAt(j));
// Recur with the next character
genPermutation(i + 1, s, used, curr, st);
// Backtrack and unmark the character
used[j] = false;
curr.deleteCharAt(curr.length() - 1);
}
}
}
static ArrayList<String> findPermutation(String s) {
// To track if a character is used
boolean[] used = new boolean[s.length()];
HashSet<String> st = new HashSet<>();
StringBuilder curr = new StringBuilder();
// Start the recursion
genPermutation(0, s, used, curr, st);
// Convert the set to a list
return new ArrayList<>(st);
}
public static void main(String[] args) {
String s = "ABC";
ArrayList<String> res = findPermutation(s);
// Print the permutations
for (String perm : res) {
System.out.print(perm + " ");
}
}
}
# Python program to print unique permutation of string using hash set.
# Function genPermutation is used to generate all possible permutation.
def genPermutation(i, s, used, curr, st):
if i == len(s):
# Add the permutation to the result set
st.add("".join(curr))
return
for j in range(len(s)):
if not used[j]:
# Mark the character as used
used[j] = True
curr.append(s[j])
# Recurse with the next character
genPermutation(i + 1, s, used, curr, st)
# Backtrack and unmark the character
used[j] = False
curr.pop()
def findPermutation(s):
# To track if a character is used
used = [False] * len(s)
st = set()
curr = []
# Start the recursion
genPermutation(0, s, used, curr, st)
# Convert the set to a list
return list(st)
if __name__ == "__main__":
s = "ABC"
res = findPermutation(s)
# Print the permutations
for perm in res:
print(perm, end=" ")
// C# program to print unique permutation of string using hash set.
using System;
using System.Collections.Generic;
class GfG {
// Function genPermutation is used to generate all possible permutation.
static void genPermutation(int i, string s, bool[] used,
List<char> curr, HashSet<string> st) {
if (i == s.Length) {
// Add the permutation to the result set
st.Add(new string(curr.ToArray()));
return;
}
for (int j = 0; j < s.Length; j++) {
if (!used[j]) {
// Mark the character as used
used[j] = true;
curr.Add(s[j]);
// Recurse with the next character
genPermutation(i + 1, s, used, curr, st);
// Backtrack and unmark the character
used[j] = false;
curr.RemoveAt(curr.Count - 1);
}
}
}
static List<string> findPermutation(string s) {
// To track if a character is used
bool[] used = new bool[s.Length];
HashSet<string> st = new HashSet<string>();
List<char> curr = new List<char>();
// Start the recursion
genPermutation(0, s, used, curr, st);
// Convert the set to a list
return new List<string>(st);
}
static void Main(string[] args) {
string s = "ABC";
List<string> res = findPermutation(s);
// Print the permutations
foreach (string perm in res) {
Console.Write(perm + " ");
}
}
}
// JavaScript program to print unique permutation of string using hash set.
// Function genPermutation is used to generate all possible permutation.
function genPermutation(i, s, used, curr, st) {
if (i === s.length) {
// Add the permutation to the result set
st.add(curr.join(''));
return;
}
for (let j = 0; j < s.length; j++) {
if (!used[j]) {
// Mark the character as used
used[j] = true;
curr.push(s[j]);
// Recurse with the next character
genPermutation(i + 1, s, used, curr, st);
// Backtrack and unmark the character
used[j] = false;
curr.pop();
}
}
}
function findPermutation(s) {
// To track if a character is used
let used = Array(s.length).fill(false);
let st = new Set();
let curr = [];
// Start the recursion
genPermutation(0, s, used, curr, st);
// Convert the set to an array
return Array.from(st);
}
// Driver Code
let s = "ABC";
let res = findPermutation(s);
// Print the permutations
console.log(res.join(' '));
OutputCBA BAC CAB BCA ABC ACB
Time Complexity: O(n*n!), where n is the size of the array.
Auxiliary Space: O(n!), space used by hash set.
[Expected Approach] Generating Only Unique Permutations
This approach is similar to generating all possible permutation. Now in this approach, while generating the permutations, at each index during recursion, we only consider the unique available characters. To track the availability of characters we use hash map.
Duplicate permutations are generated in the previous approach because it treats identical characters, like the two 'B's in "ABBC," as distinct due to their indices. For example, when placing a 'B' at an index, choosing the first or second 'B' results in the same permutation (e.g., "ABBC" or "ABBC" again), leading to duplicates.
In this approach, we prune branches in the recursive tree that would generate duplicate permutations and place only unique available characters at each index.
C++
//C++ program to print unique permutation of string
//using hash map.
#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;
// Recursive function to generate permutations
void genPermutations(int n, string &curr ,unordered_map<char, int> &cnt,
vector<string>& res) {
// Base case: If the current permutation length equals the input
// string length, add it to the result
if (curr.size() == n) {
res.push_back(curr);
return;
}
// Iterate through each character in the frequency map
for (pair<char, int> it: cnt) {
char c = it.first;
int count = it.second;
// Skip characters with a count of 0
if (count == 0)
continue;
// Include the character in the current permutation
curr.push_back(c);
// Decrease its count in the frequency map
cnt[c]--;
// Recur to build the next character in the permutation
genPermutations(n, curr, cnt , res);
// Backtrack: Remove the character and restore its count
curr.pop_back();
cnt[c]++;
}
}
// Function to find all unique permutations of the input string
vector<string> findPermutation(string s) {
// Vector to store the result
vector<string> res;
// Frequency map to count occurrences of each character
unordered_map<char, int> cnt;
// Populate the frequency map with the characters of the input string
for (char c : s)
cnt[c]++;
// To build permutations
string curr = "";
genPermutations(s.size(), curr, cnt, res);
return res;
}
int main() {
string s = "ABC";
vector<string> res = findPermutation(s);
for (string perm: res)
cout << perm << " ";
return 0;
}
//Java program to print unique permutation of string
//using hash map.
import java.util.*;
class GfG {
// Recursive function to generate permutations
static void genPermutations(int n, StringBuilder curr,
Map<Character, Integer> cnt, List<String> res) {
// Base case: If the current permutation length equals
// to the input string length, add it to the result
if (curr.length() == n) {
res.add(curr.toString());
return;
}
// Iterate through each character in the frequency map
for (Map.Entry<Character, Integer> entry : cnt.entrySet()) {
char c = entry.getKey();
int count = entry.getValue();
// Skip characters with a count of 0
if (count == 0)
continue;
// Include the character in the current permutation
curr.append(c);
// Decrease its count in the frequency map
cnt.put(c, count - 1);
// Recur to build the next character in the permutation
genPermutations(n, curr, cnt, res);
// Backtrack: Remove the character and restore its count
curr.deleteCharAt(curr.length() - 1);
cnt.put(c, count);
}
}
// Function to find all unique permutations of the input string
static ArrayList<String> findPermutation(String s) {
ArrayList<String> res = new ArrayList<>();
// Frequency map to count occurrences of each character
Map<Character, Integer> cnt = new HashMap<>();
// Populate the frequency map with the characters of the input string
for (char c : s.toCharArray())
cnt.put(c, cnt.getOrDefault(c, 0) + 1);
// To build permutations
StringBuilder curr = new StringBuilder();
genPermutations(s.length(), curr, cnt, res);
return res;
}
public static void main(String[] args) {
String s = "ABC";
List<String> res = findPermutation(s);
for (String perm : res)
System.out.print(perm + " ");
}
}
#C++ program to print unique permutation of string
#using hash map.
# Recursive function to generate permutations
def genPermutations(n, curr, cnt, res):
# Base case: If the current permutation length equals
# the input string length, add it to the result
if len(curr) == n:
res.append(curr)
return
# Iterate through each character in the frequency map
for c, count in cnt.items():
# Skip characters with a count of 0
if count == 0:
continue
# Include the character in the current permutation
cnt[c] -= 1
# Recur to build the next character in the permutation
genPermutations(n, curr + c, cnt, res)
# Backtrack: Restore the count
cnt[c] += 1
# Function to find all unique permutations of the input string
def findPermutation(s):
# List to store the result
res = []
# Frequency map to count occurrences of each character
cnt = {}
# Populate the frequency map with the characters of the input string
for c in s:
cnt[c] = cnt.get(c, 0) + 1
# Generate permutations
genPermutations(len(s), "", cnt, res)
return res
if __name__ == "__main__":
s = "ABC"
res = findPermutation(s)
print(" ".join(res))
// C# program to print unique permutation of string
// using hash map.
using System;
using System.Collections.Generic;
class GfG {
// Recursive function to generate permutations
static void genPermutations(int n, string curr,
Dictionary<char, int> cnt, List<string> res) {
// Base case: If the current permutation length equals the input
// string length, add it to the result
if (curr.Length == n) {
res.Add(curr);
return;
}
// Iterate through a copy of the dictionary's keys
foreach (char c in new List<char>(cnt.Keys)) {
int count = cnt[c];
// Skip characters with a count of 0
if (count == 0)
continue;
// Include the character in the current permutation
curr += c;
// Decrease its count in the frequency map
cnt[c]--;
// Recur to build the next character in the permutation
genPermutations(n, curr, cnt, res);
// Backtrack: Remove the character and restore its count
curr = curr.Remove(curr.Length - 1);
cnt[c]++;
}
}
// Function to find all unique permutations of the input string
static List<string> findPermutation(string s) {
// List to store the result
List<string> res = new List<string>();
// Frequency map to count occurrences of each character
Dictionary<char, int> cnt = new Dictionary<char, int>();
// Populate the frequency map with the characters of the input string
foreach (char c in s) {
if (cnt.ContainsKey(c))
cnt[c]++;
else
cnt[c] = 1;
}
// To build permutations
string curr = "";
genPermutations(s.Length, curr, cnt, res);
// Convert the result list to an array
return res;
}
static void Main(string[] args) {
string s = "ABC";
List<string> res = findPermutation(s);
foreach (string perm in res)
Console.Write(perm + " ");
}
}
//javascript program to print unique permutation of string
//using hash map.
// Recursive function to generate permutations
function genPermutations(n, curr, cnt, res) {
// Base case: If the current permutation length equals
// the input string length, add it to the result
if (curr.length === n) {
res.push(curr);
return;
}
// Iterate through each character in the frequency map
for (let c in cnt) {
if (cnt[c] === 0)
continue;
// Include the character in the current permutation
cnt[c]--;
// Recur to build the next character in the permutation
genPermutations(n, curr + c, cnt, res);
// Backtrack: Restore the count
cnt[c]++;
}
}
// Function to find all unique permutations of the input string
function findPermutation(s) {
// Array to store the result
const res = [];
// Frequency map to count occurrences of each character
const cnt = {};
// Populate the frequency map with the characters of the input string
for (const c of s)
cnt[c] = (cnt[c] || 0) + 1;
// Generate permutations
genPermutations(s.length, "", cnt, res);
return res;
}
// Driver code
const s = "ABC";
const res = findPermutation(s);
// Print the permutations
console.log(res.join(" "));
OutputCBA CAB BCA BAC ACB ABC
Time Complexity: O(n*n!), In worst case all characters were unique, so it take time equal to generating all permutations.
Auxiliary Space: O(n), used by temporary string and hash map.
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