Valid Anagram

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Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An anagram uses the exact same letters with the exact same counts, only arranged in a different order. The usual hash table approach is to count characters from one string, subtract using the other string, and make sure nothing is left unmatched.

Budibadu uses this problem to train frequency-map thinking. Once that clicks, a lot of string and hash table problems become much easier to recognise.

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

Algorithm Flow

Best Answers

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