Add solution #425

Author: JoshCrozier

README.md

@@ -413,6 +413,7 @@
 422|[Valid Word Square](./solutions/0422-valid-word-square.js)|Easy|
 423|[Reconstruct Original Digits from English](./solutions/0423-reconstruct-original-digits-from-english.js)|Medium|
 424|[Longest Repeating Character Replacement](./solutions/0424-longest-repeating-character-replacement.js)|Medium|
+425|[Word Squares](./solutions/0425-word-squares.js)|Hard|
 427|[Construct Quad Tree](./solutions/0427-construct-quad-tree.js)|Medium|
 429|[N-ary Tree Level Order Traversal](./solutions/0429-n-ary-tree-level-order-traversal.js)|Medium|
 430|[Flatten a Multilevel Doubly Linked List](./solutions/0430-flatten-a-multilevel-doubly-linked-list.js)|Medium|

solutions/0425-word-squares.js

@@ -0,0 +1,59 @@
+/**
+ * 425. Word Squares
+ * https://leetcode.com/problems/word-squares/
+ * Difficulty: Hard
+ *
+ * Given an array of unique strings words, return all the word squares you can build from
+ * words. The same word from words can be used multiple times. You can return the answer
+ * in any order.
+ *
+ * A sequence of strings forms a valid word square if the kth row and column read the same
+ * string, where 0 <= k < max(numRows, numColumns).
+ *
+ * - For example, the word sequence ["ball","area","lead","lady"] forms a word square because
+ *   each word reads the same both horizontally and vertically.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {string[][]}
+ */
+var wordSquares = function(words) {
+  const result = [];
+  const prefixMap = new Map();
+  const wordLength = words[0].length;
+
+  for (const word of words) {
+    for (let i = 0; i < word.length; i++) {
+      const prefix = word.slice(0, i);
+      if (!prefixMap.has(prefix)) {
+        prefixMap.set(prefix, []);
+      }
+      prefixMap.get(prefix).push(word);
+    }
+  }
+
+  function buildSquare(currentSquare) {
+    if (currentSquare.length === wordLength) {
+      result.push([...currentSquare]);
+      return;
+    }
+
+    const prefix = currentSquare
+      .map(word => word[currentSquare.length])
+      .join('');
+
+    const candidates = prefixMap.get(prefix) || [];
+    for (const candidate of candidates) {
+      currentSquare.push(candidate);
+      buildSquare(currentSquare);
+      currentSquare.pop();
+    }
+  }
+
+  for (const word of words) {
+    buildSquare([word]);
+  }
+
+  return result;
+};