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#127 - Word Ladder
Problem Description
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.
Solution
/**
* @param {string} beginWord
* @param {string} endWord
* @param {string[]} wordList
* @return {number}
*/
var ladderLength = function(beginWord, endWord, wordList) {
const set = new Set(wordList);
let queue = [beginWord];
let count = 1;
while (queue.length) {
const group = [];
for (const word of queue) {
if (word === endWord) {
return count;
}
for (let i = 0; i < word.length; i++) {
for (let j = 0; j < 26; j++) {
const str = word.slice(0, i) + String.fromCharCode(j + 97) + word.slice(i + 1);
if (set.has(str)) {
group.push(str);
set.delete(str);
}
}
}
}
queue = group;
count++;
}
return 0;
};