Add solution #140

Author: JoshCrozier

README.md

@@ -90,6 +90,7 @@
 137|[Single Number II](./0137-single-number-ii.js)|Medium|
 138|[Copy List with Random Pointer](./0138-copy-list-with-random-pointer.js)|Medium|
 139|[Word Break](./0139-word-break.js)|Medium|
+140|[Word Break II](./0140-word-break-ii.js)|Hard|
 141|[Linked List Cycle](./0141-linked-list-cycle.js)|Easy|
 142|[Linked List Cycle II](./0142-linked-list-cycle-ii.js)|Medium|
 143|[Reorder List](./0143-reorder-list.js)|Medium|

solutions/0140-word-break-ii.js

@@ -0,0 +1,41 @@
+/**
+ * 140. Word Break II
+ * https://leetcode.com/problems/word-break-ii/
+ * Difficulty: Hard
+ *
+ * Given a string s and a dictionary of strings wordDict, add spaces in s to
+ * construct a sentence where each word is a valid dictionary word. Return
+ * all such possible sentences in any order.
+ *
+ * Note that the same word in the dictionary may be reused multiple times in
+ * the segmentation.
+ */
+
+/**
+ * @param {string} s
+ * @param {string[]} wordDict
+ * @return {string[]}
+ */
+var wordBreak = function(s, wordDict) {
+  const result = [];
+  backtrack(result, s, wordDict, '');
+  return result;
+};
+
+function backtrack(result, s, wordDict, substr) {
+  if (!s.length) {
+    result.push(substr);
+    return;
+  }
+
+  wordDict.forEach(word => {
+    if (s.length >= word.length && word === s.substring(0, word.length)) {
+      backtrack(
+        result,
+        s.substring(word.length),
+        wordDict,
+        substr.length ? `${substr} ${word}` : word,
+      );
+    }
+  });
+}