Neetcode Day 2

click here to check out neetcode

---

242. Valid Anagram

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

My solution:

var isAnagram = function(s, t) {
    let anagram = true

    const letters = new Object;
    const letters2 = new Object;

    s.split('').forEach((letter) => {
        letters[letter]
        ? ++letters[letter]
        : letters[letter] = 1
    })

    t.split('').forEach((letter) => {
        letters2[letter]
        ? ++letters2[letter]
        : letters2[letter] = 1
    })

    Object.keys(letters).forEach((key) => {
        if (letters[key] !== letters2[key]) {
            anagram = false
        }
    })

    Object.keys(letters2).forEach((key) => {
        if (letters[key] !== letters2[key]) {
            anagram = false
        }
    })

    return anagram
};

In my solution I build two objects that contain the counts for each letter that appears in the two strings provided. Then I compare the count of each letter, looping through both objects.

I realize in retrospect that this answer is quite bloated but for the sake of honestly and transparency I wanted to include this in my blog still.

For starters, I could simply compare the lengths of the two strings. Not only would this catch edge cases early, it would also save me the trouble looping both ways at the end.

Here is a better version of my previous solution where I check the length from the start and assign the reused code to it's own funciton. A better solution but still not perfect:

var isAnagram = function(s, t) {
    if (s.length !== t.length) {
        return false
    }

    let anagram = true

    const letters = lettersCount(s)
    const letters2 = lettersCount(t)

    Object.keys(letters).forEach((key) => {
        if (letters[key] !== letters2[key]) {
            anagram = false
        }
    })

    return anagram
};

const lettersCount = (str) => {
    const lettersObj = new Object;

    str.split('').forEach((letter) => {
        lettersObj[letter]
        ? ++lettersObj[letter]
        : lettersObj[letter] = 1
    })

    return lettersObj
}

You could also sort the strings alphabetically so you need only compare each character to it's neighbor character.

---