Cracking the Code: Mastering Anagram Detection in Technical Interviews

Unlock the Secrets of Efficient Anagram Solutions with LeetCode's Insightful Challenge


4 min read

Cracking the Code: Mastering Anagram Detection in Technical Interviews

Hello there! Today, I'm excited to guide you through a popular problem that many encounter on LeetCode: determining if one string is an anagram of another. This challenge, known as the Valid Anagram (LeetCode 242) problem, is a fantastic test of your understanding of strings and hash tables.

Anagrams are words or phrases formed by rearranging the letters of a different word or phrase, using all the original letters exactly once. For instance, "listen" and "silent" are anagrams of each other.

This task might seem straightforward at first glance, but it offers a great opportunity to explore efficient algorithms and coding techniques across different programming languages. Whether you're preparing for software engineering interviews or just looking to sharpen your coding skills, mastering this problem will boost your confidence and competency.

So, let's dive into how we can solve this intriguing problem, analyze its complexity, and then implement solutions in Python, TypeScript, and Java. Stick with me, and by the end of this post, you'll be well-equipped to tackle anagram detection and similar challenges on LeetCode and beyond!

Introduction to the Problem

Consider the scenario where you're given two strings, s and t, and your goal is to discern whether t is an anagram of s.

An anagram, as defined, is a word or phrase that's formed by rearranging the letters of another, using all the original letters exactly once.

For instance, "anagram" and "nagaram" are anagrams, presenting a scenario where our function would return true.

Conversely, "rat" and "car" are not, leading to a false outcome.

Solving the Problem: A Conceptual Overview

At the heart of solving this problem is understanding how to efficiently compare the two strings to ensure they contain the same characters in any order.

The simplest approach is to sort both strings and compare them for equality. If they match, one string is indeed an anagram of the other. This method, while straightforward, carries a time complexity of O(n log n) due to the sorting operation, where n is the length of the string.

However, a more optimized solution involves using a fixed-size character count array to track the frequency of each character in both strings. By incrementing the count for each character in s and decrementing for each character in t, we ensure that if all counts return to zero, the strings are anagrams. This approach boasts a time complexity of O(n), with n being the length of the strings, significantly reducing the computational cost for larger strings.

The Solution in Python

Since this approach uses sorting it would be runtime O(n log n) where n is the length of the longer string.

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        # Sort both strings and compare
        s_sorted = sorted(s)
        t_sorted = sorted(t)
        # If sorted strings are equal, they are anagrams
        return s_sorted == t_sorted

The Solution in TypeScript

Since this approach uses sorting it would be runtime O(n log n) where n is the length of the longer string.

function isAnagram(s: string, t: string): boolean {
    // Convert strings to sorted arrays and then back to strings to compare
    return s.split("").sort().join("") === t.split("").sort().join("");

The Solution in Java

This solution does not use sorting therefore it has a runtime of O( n ). It allocates an array but the array is fixed to the length of the alphabet so the space complexity is O(1).

public class Solution {
    public boolean isAnagram(String s, String t) {
        // Create an array to count character occurrences
        int[] alphabet = new int[26];
        // Increment count for each char in s
        for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++;
        // Decrement count for each char in t
        for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--;
        // If any count is not zero, strings are not anagrams
        for (int i : alphabet) if (i != 0) return false;
        return true;


Tackling the anagram challenge not only hones your ability to manipulate strings and understand sorting algorithms but also improves your proficiency in applying efficient data structures.

As you prepare for your next technical interview, consider this problem as a stepping stone towards mastering the intricacies of algorithmic challenges. Remember, the key to excelling in software engineering interviews lies not just in solving problems but in solving them efficiently.

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