Add solution #127

Author: JoshCrozier

README.md

@@ -131,6 +131,7 @@
 123|[Best Time to Buy and Sell Stock III](./0123-best-time-to-buy-and-sell-stock-iii.js)|Hard|
 124|[Binary Tree Maximum Path Sum](./0124-binary-tree-maximum-path-sum.js)|Hard|
 125|[Valid Palindrome](./0125-valid-palindrome.js)|Easy|
+127|[Word Ladder](./0127-word-ladder.js)|Hard|
 128|[Longest Consecutive Sequence](./0128-longest-consecutive-sequence.js)|Medium|
 131|[Palindrome Partitioning](./0131-palindrome-partitioning.js)|Medium|
 133|[Clone Graph](./0133-clone-graph.js)|Medium|

solutions/0127-word-ladder.js

@@ -0,0 +1,52 @@
+/**
+ * 127. Word Ladder
+ * https://leetcode.com/problems/word-ladder/
+ * Difficulty: Hard
+ *
+ * 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.
+ */
+
+/**
+ * @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;
+};