Add solution #1046

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,118 LeetCode solutions in JavaScript
+# 1,119 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -852,6 +852,7 @@
 1042|[Flower Planting With No Adjacent](./solutions/1042-flower-planting-with-no-adjacent.js)|Medium|
 1043|[Partition Array for Maximum Sum](./solutions/1043-partition-array-for-maximum-sum.js)|Medium|
 1044|[Longest Duplicate Substring](./solutions/1044-longest-duplicate-substring.js)|Hard|
+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|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|
 1071|[Greatest Common Divisor of Strings](./solutions/1071-greatest-common-divisor-of-strings.js)|Easy|

solutions/1046-last-stone-weight.js

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+/**
+ * 1046. Last Stone Weight
+ * https://leetcode.com/problems/last-stone-weight/
+ * Difficulty: Easy
+ *
+ * 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 the heaviest two stones and
+ * smash them together. Suppose the heaviest two 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 weight of the last remaining stone. If there are no stones left, return 0.
+ */
+
+/**
+ * @param {number[]} stones
+ * @return {number}
+ */
+var lastStoneWeight = function(stones) {
+  const heap = stones.sort((a, b) => b - a);
+
+  while (heap.length > 1) {
+    const heaviest = heap.shift();
+    const secondHeaviest = heap.shift();
+    if (heaviest !== secondHeaviest) {
+      heap.push(heaviest - secondHeaviest);
+      heap.sort((a, b) => b - a);
+    }
+  }
+
+  return heap.length ? heap[0] : 0;
+};