Algorithm
Problem Name: 242. Valid Anagram
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 * 104sandtconsist of lowercase English letters.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
bool isAnagram(char* s, char* t) {
int n1[26] = { 0 }, n2[26] = { 0 };
int i, j;
while (*s && *t) {
i = *s - 'a';
j = *t - 'a';
n1[i] ++;
n2[j] ++;
s ++;
t ++;
}
if (*s || *t) return false;
for (i = 0; i < 26; i ++) {
if (n1[i] != n2[i]) return false;
}
return true;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean isAnagram(String s, String t) {
int[] counter = new int[26];
for (char c : s.toCharArray()) {
counter[c - 'a']++;
}
for (char c : t.toCharArray()) {
counter[c - 'a']--;
}
for (int i = 0; i < 26; i++) {
if (counter[i] != 0) {
return false;
}
}
return true;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const isAnagram = function(s, t) {
if (s.length !== t.length) return false;
const sh = strHash(s);
const th = strHash(t);
for (let key in sh) {
if (sh.hasOwnProperty(key) && sh[key] !== th[key]) {
return false;
}
}
return true;
};
function strHash(str) {
let res = {};
for (let i = 0; i < str.length; i++) {
if (res.hasOwnProperty(str[i])) {
res[str[i]] += 1;
} else {
res[str[i]] = 1;
}
}
return res;
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
#return sum([ord(i) for i in s])==sum([ord(j) for j in t]) and set(s)==set(t)
return sorted(s)==sorted(t)
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System.Linq;
namespace LeetCode
{
public class _0242_ValidAnagram
{
public bool IsAnagram(string s, string t)
{
if (s.Length != t.Length) return false;
var sCount = new int[26];
foreach (var ch in s)
sCount[ch - 'a']++;
var tCount = new int[26];
foreach (var ch in t)
tCount[ch - 'a']++;
return sCount.SequenceEqual(tCount);
}
}
}
Input
Output