Add solution #1049

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,120 LeetCode solutions in JavaScript
+# 1,121 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -855,6 +855,7 @@
 1046|[Last Stone Weight](./solutions/1046-last-stone-weight.js)|Easy|
 1047|[Remove All Adjacent Duplicates In String](./solutions/1047-remove-all-adjacent-duplicates-in-string.js)|Easy|
 1048|[Longest String Chain](./solutions/1048-longest-string-chain.js)|Medium|
+1049|[Last Stone Weight II](./solutions/1049-last-stone-weight-ii.js)|Medium|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|
 1071|[Greatest Common Divisor of Strings](./solutions/1071-greatest-common-divisor-of-strings.js)|Easy|
 1072|[Flip Columns For Maximum Number of Equal Rows](./solutions/1072-flip-columns-for-maximum-number-of-equal-rows.js)|Medium|

solutions/1049-last-stone-weight-ii.js

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+/**
+ * 1049. Last Stone Weight II
+ * https://leetcode.com/problems/last-stone-weight-ii/
+ * Difficulty: Medium
+ *
+ * You are given an array of integers stones where stones[i] is the weight of the ith stone.
+ *
+ * We are playing a game with the stones. On each turn, we choose any two stones and smash
+ * them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:
+ * - If x == y, both stones are destroyed, and
+ * - If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.
+ *
+ * At the end of the game, there is at most one stone left.
+ *
+ * Return the smallest possible weight of the left stone. If there are no stones left, return 0.
+ */
+
+/**
+ * @param {number[]} stones
+ * @return {number}
+ */
+var lastStoneWeightII = function(stones) {
+  const totalSum = stones.reduce((sum, weight) => sum + weight, 0);
+  const target = Math.floor(totalSum / 2);
+  const set = new Set([0]);
+
+  for (const stone of stones) {
+    const prev = new Set(set);
+    for (const sum of prev) {
+      if (sum + stone <= target) {
+        set.add(sum + stone);
+      }
+    }
+  }
+
+  return totalSum - 2 * Math.max(...set);
+};