Javascript Program To Check Whether Two Strings Are Anagram Of Each Other

Last Updated : 10 Feb, 2025

Write a function to check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an anagram of each other.

check-whether-two-strings-are-anagram-of-each-other

Method 1 (Use Sorting):

Sort both strings
Compare the sorted strings

Below is the implementation of the above idea:

JavaScript

// JavaScript program to check whether 
// two strings are anagrams of each other

// Function to check whether two strings 
// are anagram of each other 
function areAnagram(str1, str2) {
    // Get lengths of both strings
    let n1 = str1.length;
    let n2 = str2.length;

    // If length of both strings is not 
    // same, then they cannot be anagram
    if (n1 != n2)
        return false;

    // Sort both strings
    str1.sort();
    str2.sort();

    // Compare sorted strings
    for (let i = 0; i < n1; i++)
        if (str1[i] != str2[i])
            return false;

    return true;
}

// Driver Code
let str1 = ['t', 'e', 's', 't'];
let str2 = ['t', 't', 'e', 'w'];

// Function Call
if (areAnagram(str1, str2))
    console.log("The two strings are anagram of each other");
else
    console.log("The two strings are not anagram of each other");

Output:

The two strings are not anagram of each other

Time Complexity: O(nLogn)

Auxiliary space: O(1).

Method 2 (Count characters):
This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters.

Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
Compare count arrays. If both count arrays are same, then return true.

Below is the implementation of the above idea:

JavaScript

// Javascript program to check if two 
// strings are anagrams of each other
let NO_OF_CHARS = 256;

/* Function to check whether two strings
   are anagram of each other */
function areAnagram(str1, str2) {
    // Create 2 count arrays and initialize
    // all values as 0
    let count1 = new Array(NO_OF_CHARS).fill(0);
    let count2 = new Array(NO_OF_CHARS).fill(0);

    let i;

    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for (i = 0; i < str1.length && i < str2.length; i++) {
        count1[str1[i].charCodeAt(0)]++;
        count2[str2[i].charCodeAt(0)]++;
    }

    // If both strings are of different length.
    // Removing this condition will make the 
    // program fail for strings like "aaca" 
    // and "aca"
    if (str1.length != str2.length)
        return false;

    // Compare count arrays
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count1[i] != count2[i])
            return false;

    return true;
}

// Driver code
let str1 = "geeksforgeeks".split("");
let str2 = "forgeeksgeeks".split("");

// Function call
if (areAnagram(str1, str2))
    console.log("The two strings are anagram of each other");
else
    console.log("The two strings are not anagram of each other");

Output:

The two strings are anagram of each other

Time Complexity: O(n)

Auxiliary space: O(n).

Method 3 (count characters using one array):
The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are anagram of each other. Thanks to Ace for suggesting this optimization.

JavaScript

// Javascript program to check if two strings
// are anagrams of each other
let NO_OF_CHARS = 256;

// Function to check if two strings
// are anagrams of each other
function areAnagram(str1, str2) {
    // Create a count array and initialize
    // all values as 0
    let count = new Array(NO_OF_CHARS).fill(0);
    let i;

    // For each character in input strings,
    // increment count for str1 and decrement
    // count for str2 in the count array
    for (i = 0; i < str1.length; i++) {
        count[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
        count[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]--;
    }

    // If both strings are of different length,
    // they cannot be anagrams
    if (str1.length != str2.length)
        return false;

    // Check if all values in the count array are 0
    for (i = 0; i < NO_OF_CHARS; i++) {
        if (count[i] != 0) {
            return false;
        }
    }
    return true;
}

// Driver code
let str1 = "geeksforgeeks".split("");
let str2 = "forgeeksgeeks".split("");

// Function call
if (areAnagram(str1, str2))
    console.log("The two strings are anagram of each other");
else
    console.log("The two strings are not anagram of each other");

Output

The two strings are anagram of each other

Output:

The two strings are anagram of each other

Time Complexity: O(n)

Auxiliary space: O(n).

Method 4 (Using Map()):

We can optimize the space complexity of the above method by using HashMap instead of initializing 256 characters array. So in this approach, we will first count the occurrence of each unique character with the help of HashMap for the first string. Then we will reduce the count of each character while we encounter them in the second string. Finally, if the count of each character in the hash map is 0 then it means both strings are anagrams else not.

Below is the code for the above approach.

JavaScript

// JavaScript program to check if two strings are anagrams of each other

function isAnagram(a, b) {
    // Check if both strings have the same length
    if (a.length !== b.length) {
        return false;
    }

    // Create a Map to store character counts
    let map = new Map();

    // Populate the map with character counts from the first string
    for (let i = 0; i < a.length; i++) {
        map.set(a[i], (map.get(a[i]) || 0) + 1);
    }

    // Decrease character counts based on the second string
    for (let i = 0; i < b.length; i++) {
        if (!map.has(b[i])) {
            return false;
        }
        map.set(b[i], map.get(b[i]) - 1);
    }

    // Check if all counts are zero
    for (let value of map.values()) {
        if (value !== 0) {
            return false;
        }
    }

    return true;
}

// Driver code
let str1 = "geeksforgeeks";
let str2 = "forgeeksgeeks";

// Function call and console output
if (isAnagram(str1, str2)) {
    console.log("The two strings are anagrams of each other");
} else {
    console.log("The two strings are not anagrams of each other");
}

Time Complexity: O(n)

Auxiliary space: O(n) because it is using HashMap

Please suggest if someone has a better solution which is more efficient in terms of space and time.

Please refer complete article on Check whether two strings are anagram of each other for more details!

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