## Problem

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
- Input: s = "anagram", t = "nagaram"
- Output: true
- Input: s = "rat", t = "car"
- Output: false
- Constraints
- 1 <= s.length, t.length <= 5 * 104
- s and t consist of lowercase English letters.

## Solution

There are two ways of solving this problem.

### Method One

```
/**
* Since t is compared to s,
* convert all to lowercase,
* sort t and s,
* if t is exactly equal to s, then it is an anagram, else it isn't
* Runtime O(n)
*/
// const s = "anagram";
// const t = "nagaram"
const s = "rat", t = "car"
function anagram(s, t){
const sortedS = s.split('').sort().join('');
const sortedT = t.split('').sort().join('');
if (sortedS === sortedT) return true;
return false;
}
```

### Method Two

Using Map

```
/**
* Create a map object
* For each char in S, add the first occurrence with a value of 1. For subsequent occurrence, increase the value by +1.
* For each char in T, if it doesn't exist in map i.e undefined or value is <= 0, return false. If it does exist, do value -1, return true outside loop.
* @param {*} s
* @param {*} t
*/
function anagramTwo(s,t){
// for our answer to be an anagram, the word must be rearranged, hence, if the length of the two words are not same, it is not an anagram.
if(s.length !== t.length) return false;
const map = {};
for (let i of s) {
if(map[i] !== undefined) {
map[i]++
} else map[i] = 1
}
for (let i of t){
if (map[i] === undefined || map[i] <= 0) return false;
map[i]--;
}
return true
}
console.log(anagramTwo("ab", "a"))
```

If you find this helpful, like, share and bookmark. :)

## Top comments (0)