Check if two Strings are Anagrams of each other
Last Updated :
25 Jul, 2025
Given two non-empty strings s1 and s2 of lowercase letters, determine if they are anagrams — i.e., if they contain the same characters with the same frequencies.
Examples:
Input: s1 = “geeks” s2 = “kseeg”
Output: true
Explanation: Both the string have same characters with same frequency. So, they are anagrams.
Input: s1 = "allergy", s2 = "allergyy"
Output: false
Explanation: Although the characters are mostly the same, s2 contains an extra 'y' character. Since the frequency of characters differs, the strings are not anagrams.
Input: s1 = "listen", s2 = "lists"
Output: false
Explanation: The characters in the two strings are not the same — some are missing or extra. So, they are not anagrams.
[Naive Approach] Using Sorting
The idea is that if the strings are anagrams, then their characters will be the same, just rearranged. Therefore, if we sort the characters in both strings, the sorted strings will be identical if the original strings were anagrams.
We can simply sort the two given strings and compare them – if they are equal, then the original strings are anagrams of each other.
C++
#include <algorithm>
#include <iostream>
using namespace std;
bool areAnagrams(string &s1, string &s2) {
if (s1.length() != s2.length())
return false;
// Sort both strings
sort(s1.begin(), s1.end());
sort(s2.begin(), s2.end());
// Compare sorted strings
return (s1 == s2);
}
int main() {
string s1 = "geeks";
string s2 = "kseeg";
if(areAnagrams(s1, s2)){
cout << "true" << endl;
}
else{
cout << "false" << endl;
}
return 0;
}
#include <stdio.h>
#include <string.h>
// For using bool type
#include <stdbool.h>
// Function to compare two characters
// (used for sorting)
int compare(const void *a, const void *b) {
return (*(char *)a - *(char *)b);
}
// Function to check if two strings
// are anagrams
bool areAnagrams(char *s1, char *s2) {
if (strlen(s1) != strlen(s2)) return false;
// Sort both strings
qsort(s1, strlen(s1), sizeof(char), compare);
qsort(s2, strlen(s2), sizeof(char), compare);
// Compare sorted strings
return strcmp(s1, s2) == 0;
}
int main() {
char s1[] = "geeks";
char s2[] = "kseeg";
if (areAnagrams(s1, s2)) {
printf("true\n");
} else {
printf("false\n");
}
return 0;
}
import java.util.Arrays;
class GfG {
static boolean areAnagrams(String s1, String s2) {
if (s1.length() != s2.length()) return false;
// Sort both strings
char[] s1Array = s1.toCharArray();
char[] s2Array = s2.toCharArray();
Arrays.sort(s1Array);
Arrays.sort(s2Array);
// Compare sorted strings
return Arrays.equals(s1Array, s2Array);
}
public static void main(String[] args) {
String s1 = "geeks";
String s2 = "kseeg";
if(areAnagrams(s1, s2) == true){
System.out.println("true");
}
else{
System.out.println("false");
}
}
}
def areAnagrams(s1, s2):
if len(s1) != len(s2):
return False
# Sort both strings
s1 = sorted(s1)
s2 = sorted(s2)
# Compare sorted strings
return s1 == s2
if __name__ == "__main__":
s1 = "geeks"
s2 = "kseeg"
if(areAnagrams(s1,s2)):
print("true")
else:
print("false")
using System;
class GfG {
static bool areAnagrams(string s1, string s2) {
if (s1.Length != s2.Length) return false;
// Sort both strings
char[] s1Array = s1.ToCharArray();
char[] s2Array = s2.ToCharArray();
Array.Sort(s1Array);
Array.Sort(s2Array);
// Compare sorted strings
return new string(s1Array) == new string(s2Array);
}
static void Main() {
string s1 = "geeks";
string s2 = "kseeg";
if(areAnagrams(s1, s2)){
Console.WriteLine("true");
}
else{
Console.WriteLine("false");
}
}
}
function areAnagrams(s1, s2) {
if (s1.length !== s2.length) return false;
// Sort both strings
s1 = s1.split('').sort().join('');
s2 = s2.split('').sort().join('');
// Compare sorted strings
return s1 === s2;
}
// Driver Code
const s1 = "geeks";
const s2 = "kseeg";
if(areAnagrams(s1, s2)){
console.log("true");
}
else{
console.log("false");
}
Time Complexity: O(m × log(m) + n × log(n))
Auxiliary Space: In languages like C or C++, where strings can be mutable and sorted in place, the auxiliary space used is O(1). However, in languages such as Java, Python, C# and JavaScript where strings are immutable, sorting operations typically create new copies of the strings, leading to an space used is O(m + n).
[Expected Approach 1] Using Hash Map or Dictionary - O(n + m) Time and O(1) Space
The idea is to use a hash map or dictionary count the frequency of each character in both the input strings. If the frequency of every character matches in both strings, then the strings are anagrams.
Step by Step Approach:
- First, count the occurrences of each character in the first string using a HashMap.
- Then, iterate through the second string and decrement the corresponding count for each character in the same HashMap.
- After processing both strings, check the HashMap: if all character counts are zero, the strings are anagrams
- Any non-zero count indicates a mismatch in character frequency, meaning the strings are not anagrams.
C++
#include <algorithm>
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
bool areAnagrams(string &s1, string &s2) {
if(s1.size() != s2.size()){
return false;
}
// create a hashmap to store
// character frequencies
unordered_map<char, int> charCount;
// count frequency of each
// character in string s1
for(char ch: s1)
charCount[ch] += 1;
// Count frequency of each
// character in string s2
for(char ch: s2)
charCount[ch] -= 1;
// check if all frequencies are zero
for (auto& pair : charCount) {
if (pair.second != 0) {
return false;
}
}
return true;
}
int main() {
string s1 = "geeks";
string s2 = "kseeg";
if(areAnagrams(s1, s2)){
cout << "true";
}
else{
cout << "false";
}
return 0;
}
import java.util.HashMap;
class GfG {
static boolean areAnagrams(String s1, String s2) {
if(s1.length() != s2.length()){
return false;
}
// create a hashmap to store
// character frequencies
HashMap<Character, Integer> charCount =
new HashMap<>();
// count frequency of each
// character in string s1
for (char ch : s1.toCharArray())
charCount.put(ch,
charCount.getOrDefault(ch, 0) + 1);
// count frequency of each
// character in string s2
for (char ch : s2.toCharArray())
charCount.put(ch,
charCount.getOrDefault(ch, 0) - 1);
// check if all frequencies are zero
for (var pair : charCount.entrySet()) {
if (pair.getValue() != 0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
String s1 = "geeks";
String s2 = "kseeg";
System.out.println(areAnagrams(s1, s2) ? "true" : "false");
}
}
def areAnagrams(s1, s2):
if len(s1) != len(s2):
return False
# create a hashmap to store
# character frequencies
charCount = {}
# count frequency of each
# character in string s1
for ch in s1:
charCount[ch] = charCount.get(ch, 0) + 1
# count frequency of each
# character in string s2
for ch in s2:
charCount[ch] = charCount.get(ch, 0) - 1
# check if all frequencies are zero
for value in charCount.values():
if value != 0:
return False
return True
if __name__ == "__main__":
s1 = "geeks"
s2 = "kseeg"
if areAnagrams(s1, s2):
print("true")
else:
print("false")
using System;
using System.Collections.Generic;
class GfG {
static bool areAnagrams(string s1, string s2) {
if (s1.Length != s2.Length) return false;
// create a dictionary to store
// character frequencies
Dictionary<char, int> charCount =
new Dictionary<char, int>();
// count frequency of each
// character in string s1
foreach (char ch in s1)
charCount[ch] =
charCount.GetValueOrDefault(ch, 0) + 1;
// count frequency of each character in string s2
foreach (char ch in s2)
charCount[ch] =
charCount.GetValueOrDefault(ch, 0) - 1;
// check if all frequencies are zero
foreach (var pair in charCount) {
if (pair.Value != 0)
return false;
}
return true;
}
static void Main(string[] args) {
string s1 = "geeks";
string s2 = "kseeg";
if(areAnagrams(s1, s2)){
Console.WriteLine("true");
}
else{
Console.WriteLine("false");
}
}
}
function areAnagrams(s1, s2) {
if (s1.length !== s2.length) return false;
// create a hashmap to store
// character frequencies
const charCount = {};
// count frequency of each
// character in string s1
for (let ch of s1)
charCount[ch] = (charCount[ch] || 0) + 1;
// count frequency of each
// character in string s2
for (let ch of s2)
charCount[ch] = (charCount[ch] || 0) - 1;
// check if all frequencies are zero
for (let key in charCount) {
if (charCount[key] !== 0) {
return false;
}
}
return true;
}
// Driver Code
const s1 = "geeks";
const s2 = "kseeg";
if(areAnagrams(s1, s2)){
console.log("true");
}
else{
console.log("false");
}
[Expected Approach 2] Using Frequency Array - O(n + m) Time and O(1) Space
Instead of using a hash map to store the frequency of each character, we can create a frequency array of size 26 by using characters as index in this array. The frequency of ‘a’ is going to be stored at index 0, ‘b’ at 1, and so on. To find the index of a character, we subtract character a’s ASCII value from the ASCII value of the character.
Count character frequency in the first string, then for each character in the second string, decrement its count from the frequency array. If the strings are anagrams, all positions in the frequency array will be zero. Any non-zero position means the frequency of that character is not equal in both the strings.
Working:
C++
#include <iostream>
#include <vector>
using namespace std;
bool areAnagrams(string &s1, string &s2) {
if (s1.length() != s2.length()) return false;
// for lowercase a-z
vector<int> freq(26, 0);
// Count frequency of each character in s1
for(char ch : s1)
freq[ch - 'a']++;
// Subtract frequency using characters from s2
for(char ch : s2)
freq[ch - 'a']--;
// If any count is not zero, not an anagram
for(int count : freq) {
if (count != 0)
return false;
}
return true;
}
int main() {
string s1 = "geeks";
string s2 = "kseeg";
if (areAnagrams(s1, s2))
cout << "true" << endl;
else
cout << "false" << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
bool areAnagrams(char *s1, char *s2) {
if (strlen(s1) != strlen(s2)) return false;
// for lowercase a-z
int freq[26] = {0};
// Count frequency of each character in s1
for (int i = 0; s1[i] != '\0'; i++)
freq[s1[i] - 'a']++;
// Subtract frequency using characters from s2
for (int i = 0; s2[i] != '\0'; i++)
freq[s2[i] - 'a']--;
// Check if all frequencies are zero
for (int i = 0; i < 26; i++) {
if (freq[i] != 0)
return false;
}
return true;
}
int main() {
char s1[] = "geeks";
char s2[] = "kseeg";
if (areAnagrams(s1, s2))
printf("true\n");
else
printf("false\n");
return 0;
}
class GfG {
static boolean areAnagrams(String s1, String s2) {
if (s1.length() != s2.length()) return false;
// for lowercase a-z
int[] freq = new int[26];
// Count frequency of each character in s1
for (int i = 0; i < s1.length(); i++)
freq[s1.charAt(i) - 'a']++;
// Subtract frequency using characters from s2
for (int i = 0; i < s2.length(); i++)
freq[s2.charAt(i) - 'a']--;
// Check if all frequencies are zero
for (int count : freq) {
if (count != 0)
return false;
}
return true;
}
public static void main(String[] args) {
String s1 = "geeks";
String s2 = "kseeg";
if (areAnagrams(s1, s2))
System.out.println("true");
else
System.out.println("false");
}
}
def areAnagrams(s1, s2):
if len(s1) != len(s2):
return False
# for lowercase a-z
freq = [0] * 26
# Count frequency of each character in s1
for ch in s1:
freq[ord(ch) - ord('a')] += 1
# Subtract frequency using characters from s2
for ch in s2:
freq[ord(ch) - ord('a')] -= 1
# Check if all frequencies are zero
for count in freq:
if count != 0:
return False
return True
if __name__ == "__main__":
s1 = "geeks"
s2 = "kseeg"
if areAnagrams(s1, s2):
print("true")
else:
print("false")
using System;
class GfG {
static bool areAnagrams(string s1, string s2) {
if (s1.Length != s2.Length) return false;
// for lowercase a-z
int[] freq = new int[26];
// Count frequency of each character in s1
foreach (char ch in s1)
freq[ch - 'a']++;
// Subtract frequency using characters from s2
foreach (char ch in s2)
freq[ch - 'a']--;
// Check if all frequencies are zero
foreach (int count in freq) {
if (count != 0)
return false;
}
return true;
}
static void Main() {
string s1 = "geeks";
string s2 = "kseeg";
if (areAnagrams(s1, s2))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
function areAnagrams(s1, s2) {
if (s1.length !== s2.length) return false;
// for lowercase a-z
let freq = new Array(26).fill(0);
// Count frequency of each character in s1
for (let ch of s1)
freq[ch.charCodeAt(0) - 'a'.charCodeAt(0)]++;
// Subtract frequency using characters from s2
for (let ch of s2)
freq[ch.charCodeAt(0) - 'a'.charCodeAt(0)]--;
// Check if all frequencies are zero
for (let count of freq) {
if (count !== 0)
return false;
}
return true;
}
// Driver Code
const s1 = "geeks";
const s2 = "kseeg";
if (areAnagrams(s1, s2)) {
console.log("true");
} else {
console.log("false");
}
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