Valid Anagram

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This challenge is like checking if you can spell one word using exact letters from another. In Valid Anagram, you’re given two strings, s and t. Return true if t is an anagram of s. You must use every letter from the original word exactly once, just rearranged.

The "secret sauce" is a Frequency Map. First, verify if the two strings are the same length—if not, you can stop! Then, count character appearances in the first string. As you walk through the second, subtract from those counts. If every tally is exactly zero, you have a perfect match! This approach is efficient, giving linear time complexity O(n). It’s the gold standard for string comparison where order doesn't matter, but totals do!

For text and token tasks, be precise with index movement and substring boundaries. Most hidden failures come from partial-match handling and boundary cuts, so keep comparisons explicit and avoid assumptions about implicit separators or formatting not guaranteed by the input contract.

Examples

Example 1

Input

s = "anagram", t = "nagaram"

Output

true

Explanation

Both strings contain the same letters with the same counts.

Example 2

Input

s = "rat", t = "car"

Output

false

Explanation

The letters do not match.

Example 3

Input

s = "aacc", t = "ccac"

Output

false

Explanation

The character counts are different.

Algorithm Flow

Recommendation Algorithm Flow for Valid Anagram - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public boolean valid_anagram(String s, String t) {
        if (s.length() != t.length()) return false;
        Map<Character, Integer> count = new HashMap<>();
        for (char ch : s.toCharArray()) count.put(ch, count.getOrDefault(ch, 0) + 1);
        for (char ch : t.toCharArray()) {
            if (!count.containsKey(ch) || count.get(ch) == 0) return false;
            count.put(ch, count.get(ch) - 1);
        }
        return true;
    }
}

javascript

function valid_anagram(s, t) {
    if (s.length !== t.length) return false;
    const count = {};
    for (const ch of s) count[ch] = (count[ch] || 0) + 1;
    for (const ch of t) {
        if (!count[ch]) return false;
        count[ch]--;
    }
    return true;
}

php

<?php
function valid_anagram($s, $t) {
    if (strlen($s) !== strlen($t)) return false;
    $count = [];
    foreach (str_split($s) as $ch) {
        $count[$ch] = ($count[$ch] ?? 0) + 1;
    }
    foreach (str_split($t) as $ch) {
        if (!isset($count[$ch]) || $count[$ch] === 0) return false;
        $count[$ch]--;
    }
    return true;
}
?>

python

def valid_anagram(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
    count = {}
    for ch in s:
        count[ch] = count.get(ch, 0) + 1
    for ch in t:
        if ch not in count:
            return False
        count[ch] -= 1
        if count[ch] == 0:
            del count[ch]
    return len(count) == 0

rust

use std::collections::HashMap;
fn valid_anagram(s: String, t: String) -> bool {
    if s.len() != t.len() {
        return false;
    }
    let mut count: HashMap<char, i32> = HashMap::new();
    for ch in s.chars() {
        *count.entry(ch).or_insert(0) += 1;
    }
    for ch in t.chars() {
        match count.get_mut(&ch) {
            Some(value) if *value > 0 => *value -= 1,
            _ => return false,
        }
    }
    true
}

typescript

function valid_anagram(s: string, t: string): boolean {
    if (s.length !== t.length) return false;
    var count: { [key: string]: number } = {};
    var i: number;
    for (i = 0; i < s.length; i++) {
        var a = s.charAt(i);
        count[a] = (count[a] || 0) + 1;
    }
    for (i = 0; i < t.length; i++) {
        var b = t.charAt(i);
        if (!count[b]) return false;
        count[b]--;
    }
    return true;
}

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Valid Anagram

Examples

Algorithm Flow

Best Answers

Solutions Coming Soon

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