242. Valid Anagram

At a Glance

Topic: arrays-hashing
Pattern: Hash Map Counting
Difficulty: Easy
Companies: Amazon, Google, Meta, Bloomberg, Microsoft
Frequency: High
LeetCode: 242

Problem (one-liner)

Given two strings first and second, return true if second is an anagram of first (same multiset of characters), else false. Typically lowercase English letters.

Recognition Cues

"Anagram", "permutation of letters", "same characters same counts"
Length mismatch immediately implies false

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

Loading diagram…

Approaches

Brute force — sort both strings and compare — O(n log n) time / O(1) extra if sorting in place allowed on copies.
Better — two frequency maps merge — O(n) time / O(1) space (alphabet size fixed).
Optimal — single array of 26 counts: increment for first, decrement for second — O(n) time / O(1) space.

Optimal Solution

Go

package main

func isAnagram(first string, second string) bool {
	if len(first) != len(second) {
		return false
	}
	counts := [26]int{}
	for index := range first {
		counts[first[index]-'a']++
		counts[second[index]-'a']--
	}
	for _, value := range counts {
		if value != 0 {
			return false
		}
	}
	return true
}

JavaScript

function isAnagram(first, second) {
	if (first.length !== second.length) {
		return false;
	}
	const counts = new Array(26).fill(0);
	for (let index = 0; index < first.length; index++) {
		counts[first.charCodeAt(index) - 97]++;
		counts[second.charCodeAt(index) - 97]--;
	}
	return counts.every((value) => value === 0);
}