Add solution #139

Author: JoshCrozier

README.md

@@ -87,6 +87,7 @@
 136|[Single Number](./0136-single-number.js)|Easy|
 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|
 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/0139-word-break.js

@@ -0,0 +1,30 @@
+/**
+ * 139. Word Break
+ * https://leetcode.com/problems/word-break/
+ * Difficulty: Medium
+ *
+ * Given a string s and a dictionary of strings wordDict, return true if s can be segmented into
+ * a space-separated sequence of one or more dictionary words.
+ *
+ * Note that the same word in the dictionary may be reused multiple times in the segmentation.
+ */
+
+/**
+ * @param {string} s
+ * @param {string[]} wordDict
+ * @return {boolean}
+ */
+var wordBreak = function(s, wordDict) {
+  const result = [1, ...new Array(s.length + 1).fill(0)];
+
+  for (let i = 1; i <= s.length; i++) {
+    for (let j = 0; j < i; j++) {
+      if (result[j] && wordDict.includes(s.slice(j, i))) {
+        result[i] = 1;
+        break;
+      }
+    }
+  }
+
+  return result[s.length];
+};