# 242. Valid Anagram

## Description

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

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

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


Example 2:

Input: s = "rat", t = "car"
Output: false


Constraints:

• 1 <= s.length, t.length <= 5 * 104
• s and t consist of lowercase English letters.

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

## Solutions

Solution 1: Counting

We first determine whether the length of the two strings is equal. If they are not equal, the characters in the two strings must be different, so return false.

Otherwise, we use a hash table or an array of length $26$ to record the number of times each character appears in the string $s$, and then traverse the other string $t$. Each time we traverse a character, we subtract the number of times the corresponding character appears in the hash table by one. If the number of times after subtraction is less than $0$, the number of times the character appears in the two strings is different, return false. If after traversing the two strings, all the character counts in the hash table are $0$, it means that the characters in the two strings appear the same number of times, return true.

The time complexity is $O(n)$, the space complexity is $O(C)$, where $n$ is the length of the string; and $C$ is the size of the character set, which is $C=26$ in this problem.

• class Solution {
public boolean isAnagram(String s, String t) {
if (s.length() != t.length()) {
return false;
}
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
--cnt[t.charAt(i) - 'a'];
}
for (int i = 0; i < 26; ++i) {
if (cnt[i] != 0) {
return false;
}
}
return true;
}
}

• class Solution {
public:
bool isAnagram(string s, string t) {
if (s.size() != t.size()) {
return false;
}
vector<int> cnt(26);
for (int i = 0; i < s.size(); ++i) {
++cnt[s[i] - 'a'];
--cnt[t[i] - 'a'];
}
return all_of(cnt.begin(), cnt.end(), [](int x) { return x == 0; });
}
};

• class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False
cnt = Counter(s)
for c in t:
cnt[c] -= 1
if cnt[c] < 0:
return False
return True


• func isAnagram(s string, t string) bool {
if len(s) != len(t) {
return false
}
cnt := [26]int{}
for i := 0; i < len(s); i++ {
cnt[s[i]-'a']++
cnt[t[i]-'a']--
}
for _, v := range cnt {
if v != 0 {
return false
}
}
return true
}

• function isAnagram(s: string, t: string): boolean {
if (s.length !== t.length) {
return false;
}
const cnt = new Array(26).fill(0);
for (let i = 0; i < s.length; ++i) {
++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
--cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
}
return cnt.every(x => x === 0);
}


• /**
* @param {string} s
* @param {string} t
* @return {boolean}
*/
var isAnagram = function (s, t) {
if (s.length !== t.length) {
return false;
}
const cnt = new Array(26).fill(0);
for (let i = 0; i < s.length; ++i) {
++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
--cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
}
return cnt.every(x => x === 0);
};


• public class Solution {
public bool IsAnagram(string s, string t) {
if (s.Length != t.Length) {
return false;
}
int[] cnt = new int[26];
for (int i = 0; i < s.Length; ++i) {
++cnt[s[i] - 'a'];
--cnt[t[i] - 'a'];
}
return cnt.All(x => x == 0);
}
}

• impl Solution {
pub fn is_anagram(s: String, t: String) -> bool {
let n = s.len();
let m = t.len();
if n != m {
return false;
}
let (s, t) = (s.as_bytes(), t.as_bytes());
let mut count = [0; 26];
for i in 0..n {
count[(s[i] - b'a') as usize] += 1;
count[(t[i] - b'a') as usize] -= 1;
}
count.iter().all(|&c| c == 0)
}
}