Add solution #244

JoshCrozier · JoshCrozier · commit ea7c0cde4e28 · 2025-06-08T18:56:54.000-05:00
diff --git a/README.md b/README.md
@@ -233,6 +233,7 @@
 241|[Different Ways to Add Parentheses](./solutions/0241-different-ways-to-add-parentheses.js)|Medium|
 242|[Valid Anagram](./solutions/0242-valid-anagram.js)|Easy|
 243|[Shortest Word Distance](./solutions/0243-shortest-word-distance.js)|Easy|
+244|[Shortest Word Distance II](./solutions/0244-shortest-word-distance-ii.js)|Medium|
 257|[Binary Tree Paths](./solutions/0257-binary-tree-paths.js)|Easy|
 258|[Add Digits](./solutions/0258-add-digits.js)|Easy|
 260|[Single Number III](./solutions/0260-single-number-iii.js)|Medium|
diff --git a/solutions/0244-shortest-word-distance-ii.js b/solutions/0244-shortest-word-distance-ii.js
@@ -0,0 +1,50 @@
+/**
+ * 244. Shortest Word Distance II
+ * https://leetcode.com/problems/shortest-word-distance-ii/
+ * Difficulty: Medium
+ *
+ * Design a data structure that will be initialized with a string array, and then it should
+ * answer queries of the shortest distance between two different strings from the array.
+ *
+ * Implement the WordDistance class:
+ * - WordDistance(String[] wordsDict) initializes the object with the strings array wordsDict.
+ * - int shortest(String word1, String word2) returns the shortest distance between word1 and
+ *   word2 in the array wordsDict.
+ */
+
+/**
+ * @param {string[]} wordsDict
+ */
+var WordDistance = function(wordsDict) {
+  this.wordIndices = new Map();
+  for (let i = 0; i < wordsDict.length; i++) {
+    if (!this.wordIndices.has(wordsDict[i])) {
+      this.wordIndices.set(wordsDict[i], []);
+    }
+    this.wordIndices.get(wordsDict[i]).push(i);
+  }
+};
+
+/**
+ * @param {string} word1
+ * @param {string} word2
+ * @return {number}
+ */
+WordDistance.prototype.shortest = function(word1, word2) {
+  const indices1 = this.wordIndices.get(word1);
+  const indices2 = this.wordIndices.get(word2);
+  let minDistance = Infinity;
+  let i = 0;
+  let j = 0;
+
+  while (i < indices1.length && j < indices2.length) {
+    minDistance = Math.min(minDistance, Math.abs(indices1[i] - indices2[j]));
+    if (indices1[i] < indices2[j]) {
+      i++;
+    } else {
+      j++;
+    }
+  }
+
+  return minDistance;
+};
