Description
Given two strings s and t, return true if t is an anagram of s, and false otherwise.
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 * 104s 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.
PythonJavaC++GoTypeScriptRustJavaScriptC#C
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(code-box)
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;
}
}(code-box)
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; });
}
};(code-box)
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
}(code-box)
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);
}(code-box)
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 mut s = s.chars().collect::<Vec<char>>();
let mut t = t.chars().collect::<Vec<char>>();
s.sort();
t.sort();
for i in 0..n {
if s[i] != t[i] {
return false;
}
}
true
}
}(code-box)
/**
* @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);
};(code-box)
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);
}
}(code-box)
int cmp(const void* a, const void* b) {
return *(char*) a - *(char*) b;
}
bool isAnagram(char* s, char* t) {
int n = strlen(s);
int m = strlen(t);
if (n != m) {
return 0;
}
qsort(s, n, sizeof(char), cmp);
qsort(t, n, sizeof(char), cmp);
return !strcmp(s, t);
}(code-box)
Solution 2
PythonRustC
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
return Counter(s) == Counter(t)(code-box)
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)
}
}(code-box)
bool isAnagram(char* s, char* t) {
int n = strlen(s);
int m = strlen(t);
if (n != m) {
return 0;
}
int count[26] = {0};
for (int i = 0; i < n; i++) {
count[s[i] - 'a']++;
count[t[i] - 'a']--;
}
for (int i = 0; i < 26; i++) {
if (count[i]) {
return 0;
}
}
return 1;
}(code-box)
