Word Ladder

Problem statement

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]Output: 5Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]Output: 0Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord, endWord, and wordList[i] consist of lowercase English letters.
• beginWord != endWord
• All the words in wordList are unique.

My solution

/**
 * @param {string} beginWord
 * @param {string} endWord
 * @param {string[]} wordList
 * @return {number}
 */
var ladderLength = function(beginWord, endWord, wordList) {
    const seen = new Set();
    const dict= new Set(wordList);
    let len = 1;
    
    const queue = [beginWord]
    
    while(queue.length > 0) {
        const N = queue.length;
        for (let w = 0; w < N; w++) {
            const word = queue.shift();
            if(word === endWord) {
                return len;
            }
            const arr = word.split("");
            for (let i = 0; i < arr.length; i++) {
                for (j = 0; j < 26; j++) {
                    arr[i] = String.fromCharCode(97 + j)
                    const newWord = arr.join("")
                    if (!seen.has(newWord) && dict.has(newWord)) {
                        seen.add(newWord)
                        queue.push(newWord)
                    }
                }
                // Reset word
                arr[i] = word[i]
            }
        }
        len++
    }
    
    return 0;
};