LeetCode Challenge: 242. Valid Anagram - JavaScript Solution 🚀

Top Interview 150

The Valid Anagram problem is a straightforward challenge involving string manipulation and character counting. Let’s solve LeetCode 242: Valid Anagram step by step.

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🚀 Problem Description

Given two strings s and t:

• Return true if t is an anagram of s.
• Otherwise, return false.

An anagram is a word or phrase formed by rearranging the letters of another. Both strings must have the same characters in the same frequencies.

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💡 Examples

Example 1

Input: s = "anagram", t = "nagaram"  
Output: true  
Explanation: Both strings have the same characters with the same frequencies.

Example 2

Input: s = "rat", t = "car"  
Output: false  
Explanation: The characters in `t` are not the same as in `s`.

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🏆 JavaScript Solution

We can efficiently solve this problem by counting the frequency of characters in both strings and comparing the counts.

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Implementation

var isAnagram = function(s, t) {
    if (s.length !== t.length) return false;

    const charCount = {};

    for (let char of s) {
        charCount[char] = (charCount[char] || 0) + 1;
    }

    for (let char of t) {
        if (!charCount[char]) return false;
        charCount[char]--;
    }

    return true;
};

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🔍 How It Works

1. Count Characters in s:

   • Use a hash map (charCount) to track the frequency of each character in s.

2. Subtract Characters in t:

   • For each character in t, decrement its count in charCount.
   • If a character in t is missing or its count becomes negative, return false.

3. Return True:

   • If all characters match and their counts balance, return true.

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🔑 Complexity Analysis

• Time Complexity: O(n), where n is the length of s (or t).

  • Each character is processed once.

• Space Complexity: O(1), as the hash map stores at most 26 entries for lowercase English letters.

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📋 Dry Run

Input: s = "anagram", t = "nagaram"

Output: true

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Follow-Up: Unicode Characters
For inputs containing Unicode characters, the solution can be adapted as follows:

• Instead of assuming lowercase English letters, allow any character as a key in the hash map.
• This approach is already supported by the above implementation, as JavaScript hash maps ({} or Map) can store any string as a key.

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✨ Pro Tips for Interviews

1. Discuss Sorting Approach:
   • An alternative solution involves sorting both strings and comparing them.
   • Time Complexity: O(nlogn) due to sorting.

var isAnagram = function(s, t) {
    return s.split("").sort().join("") === t.split("").sort().join("");
};

`

1. Clarify Edge Cases:
   • Different lengths: s = "abc", t = "abcd".
   • Empty strings: s = "", t = "".

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📚 Learn More

Check out the full explanation and code walkthrough on my previous Dev.to post:
👉 Word Pattern - JavaScript Solution

What’s your preferred method to solve this problem? Let’s discuss! 🚀

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Comments

[Rahul Kumar Barnwal]

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