Word Ladder

hardcodingAlgorithmsBFSGraphStrings~45 min

Asked at Google, Meta

Overview

Given beginWord, endWord, and a word list, find the shortest transformation sequence length from beginWord to endWord, where:

Only one letter can be changed at a time.
Each transformed word must exist in the word list.

Return 0 if no such sequence exists.

Constraints

1 <= beginWord.length <= 10
All words are the same length
1 <= wordList.length <= 5000

Examples

ladderLength("hit", "cog", ["hot","dot","dog","lot","log","cog"]);
// => 5 (hit → hot → dot → dog → cog)

ladderLength("hit", "cog", ["hot","dot","dog","lot","log"]);
// => 0 (no path to "cog")

Solution

Reveal solution

function ladderLength(beginWord, endWord, wordList) {
  const wordSet = new Set(wordList);
  if (!wordSet.has(endWord)) return 0;
  const queue = [[beginWord, 1]];
  const visited = new Set([beginWord]);
  while (queue.length) {
    const [word, steps] = queue.shift();
    for (let i = 0; i < word.length; i++) {
      for (let c = 97; c <= 122; c++) {
        const next = word.slice(0, i) + String.fromCharCode(c) + word.slice(i + 1);
        if (next === endWord) return steps + 1;
        if (wordSet.has(next) && !visited.has(next)) {
          visited.add(next);
          queue.push([next, steps + 1]);
        }
      }
    }
  }
  return 0;
}

Resources

Word Ladder — LeetCode

word-ladder.js

Word Ladder

hardcodingAlgorithmsBFS

Overview

Given beginWord, endWord, and a word list, find the shortest transformation sequence length from beginWord to endWord, where:

Only one letter can be changed at a time.
Each transformed word must exist in the word list.

Return 0 if no such sequence exists.

Constraints

1 <= beginWord.length <= 10
All words are the same length
1 <= wordList.length <= 5000

Examples

ladderLength("hit", "cog", ["hot","dot","dog","lot","log","cog"]);
// => 5 (hit → hot → dot → dog → cog)

ladderLength("hit", "cog", ["hot","dot","dog","lot","log"]);
// => 0 (no path to "cog")

Solution

Reveal solution

function ladderLength(beginWord, endWord, wordList) {
  const wordSet = new Set(wordList);
  if (!wordSet.has(endWord)) return 0;
  const queue = [[beginWord, 1]];
  const visited = new Set([beginWord]);
  while (queue.length) {
    const [word, steps] = queue.shift();
    for (let i = 0; i < word.length; i++) {
      for (let c = 97; c <= 122; c++) {
        const next = word.slice(0, i) + String.fromCharCode(c) + word.slice(i + 1);
        if (next === endWord) return steps + 1;
        if (wordSet.has(next) && !visited.has(next)) {
          visited.add(next);
          queue.push([next, steps + 1]);
        }
      }
    }
  }
  return 0;
}

Resources

Word Ladder — LeetCode

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