feat: add ts solution to lc problem: No.2029

Author: rain84

solution/2000-2099/2029.Stone Game IX/README.md

@@ -42,7 +42,7 @@ tags:
 <strong>输出：</strong>true
 <strong>解释：</strong>游戏进行如下：
 - 回合 1：Alice 可以移除任意一个石子。
-- 回合 2：Bob 移除剩下的石子。 
+- 回合 2：Bob 移除剩下的石子。
 已移除的石子的值总和为 1 + 2 = 3 且可以被 3 整除。因此，Bob 输，Alice 获胜。
 </pre>
 
@@ -51,7 +51,7 @@ tags:
 <pre>
 <strong>输入：</strong>stones = [2]
 <strong>输出：</strong>false
-<strong>解释：</strong>Alice 会移除唯一一个石子，已移除石子的值总和为 2 。 
+<strong>解释：</strong>Alice 会移除唯一一个石子，已移除石子的值总和为 2 。
 由于所有石子都已移除，且值总和无法被 3 整除，Bob 获胜。
 </pre>
 
@@ -229,4 +229,42 @@ function stoneGameIX(stones: number[]): boolean {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 2: Simulation
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function stoneGameIX(stones: number[]): boolean {
+    if (stones.length === 1) return false;
+
+    const cnt = Array(3).fill(0);
+    for (const x of stones) cnt[x % 3]++;
+
+    const check = (x: number, cnt: number[]): boolean => {
+        let c = 1;
+        if (--cnt[x] < 0) return false;
+
+        while (cnt[1] || cnt[2]) {
+            if (cnt[x]) {
+                cnt[x]--;
+                x = x === 1 ? 2 : 1;
+            } else return (c + cnt[0]) % 2 === 1;
+            c++;
+        }
+
+        return false;
+    };
+
+    return check(1, [...cnt]) || check(2, [...cnt]);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/2000-2099/2029.Stone Game IX/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <strong>Output:</strong> true
 <strong>Explanation:</strong>&nbsp;The game will be played as follows:
 - Turn 1: Alice can remove either stone.
-- Turn 2: Bob removes the remaining stone. 
+- Turn 2: Bob removes the remaining stone.
 The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.
 </pre>
 
@@ -45,7 +45,7 @@ The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob
 <pre>
 <strong>Input:</strong> stones = [2]
 <strong>Output:</strong> false
-<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2. 
+<strong>Explanation:</strong>&nbsp;Alice will remove the only stone, and the sum of the values on the removed stones is 2.
 Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.
 </pre>
 
@@ -222,4 +222,42 @@ function stoneGameIX(stones: number[]): boolean {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 2: Simulation
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function stoneGameIX(stones: number[]): boolean {
+    if (stones.length === 1) return false;
+
+    const cnt = Array(3).fill(0);
+    for (const x of stones) cnt[x % 3]++;
+
+    const check = (x: number, cnt: number[]): boolean => {
+        let c = 1;
+        if (--cnt[x] < 0) return false;
+
+        while (cnt[1] || cnt[2]) {
+            if (cnt[x]) {
+                cnt[x]--;
+                x = x === 1 ? 2 : 1;
+            } else return (c + cnt[0]) % 2 === 1;
+            c++;
+        }
+
+        return false;
+    };
+
+    return check(1, [...cnt]) || check(2, [...cnt]);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/2000-2099/2029.Stone Game IX/Solution2.ts

@@ -0,0 +1,23 @@
+function stoneGameIX(stones: number[]): boolean {
+    if (stones.length === 1) return false;
+
+    const cnt = Array(3).fill(0);
+    for (const x of stones) cnt[x % 3]++;
+
+    const check = (x: number, cnt: number[]): boolean => {
+        let c = 1;
+        if (--cnt[x] < 0) return false;
+
+        while (cnt[1] || cnt[2]) {
+            if (cnt[x]) {
+                cnt[x]--;
+                x = x === 1 ? 2 : 1;
+            } else return (c + cnt[0]) % 2 === 1;
+            c++;
+        }
+
+        return false;
+    };
+
+    return check(1, [...cnt]) || check(2, [...cnt]);
+}