Add solution #2029

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,696 LeetCode solutions in JavaScript
+# 1,697 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1556,6 +1556,7 @@
 2025|[Maximum Number of Ways to Partition an Array](./solutions/2025-maximum-number-of-ways-to-partition-an-array.js)|Hard|
 2027|[Minimum Moves to Convert String](./solutions/2027-minimum-moves-to-convert-string.js)|Easy|
 2028|[Find Missing Observations](./solutions/2028-find-missing-observations.js)|Medium|
+2029|[Stone Game IX](./solutions/2029-stone-game-ix.js)|Medium|
 2033|[Minimum Operations to Make a Uni-Value Grid](./solutions/2033-minimum-operations-to-make-a-uni-value-grid.js)|Medium|
 2037|[Minimum Number of Moves to Seat Everyone](./solutions/2037-minimum-number-of-moves-to-seat-everyone.js)|Easy|
 2047|[Number of Valid Words in a Sentence](./solutions/2047-number-of-valid-words-in-a-sentence.js)|Easy|

solutions/2029-stone-game-ix.js

@@ -0,0 +1,33 @@
+/**
+ * 2029. Stone Game IX
+ * https://leetcode.com/problems/stone-game-ix/
+ * Difficulty: Medium
+ *
+ * Alice and Bob continue their games with stones. There is a row of n stones, and each stone has
+ * an associated value. You are given an integer array stones, where stones[i] is the value of the
+ * ith stone.
+ *
+ * Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any
+ * stone from stones. The player who removes a stone loses if the sum of the values of all
+ * removed stones is divisible by 3. Bob will win automatically if there are no remaining stones
+ * (even if it is Alice's turn).
+ *
+ * Assuming both players play optimally, return true if Alice wins and false if Bob wins.
+ */
+
+/**
+ * @param {number[]} stones
+ * @return {boolean}
+ */
+var stoneGameIX = function(stones) {
+  const remainderCount = [0, 0, 0];
+  for (const stone of stones) {
+    remainderCount[stone % 3]++;
+  }
+
+  if (remainderCount[0] % 2 === 0) {
+    return remainderCount[1] >= 1 && remainderCount[2] >= 1;
+  }
+
+  return Math.abs(remainderCount[1] - remainderCount[2]) > 2;
+};