{ "data": { "question": { "questionId": "1346", "questionFrontendId": "2202", "categoryTitle": "Algorithms", "boundTopicId": 176031, "title": "Maximize the Topmost Element After K Moves", "titleSlug": "maximize-the-topmost-element-after-k-moves", "content": "

You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0] is the topmost element of the pile.

\n\n

In one move, you can perform either of the following:

\n\n\n\n

You are also given an integer k, which denotes the total number of moves to be made.

\n\n

Return the maximum value of the topmost element of the pile possible after exactly k moves. In case it is not possible to obtain a non-empty pile after k moves, return -1.

\n\n

 

\n

Example 1:

\n\n
\nInput: nums = [5,2,2,4,0,6], k = 4\nOutput: 5\nExplanation:\nOne of the ways we can end with 5 at the top of the pile after 4 moves is as follows:\n- Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6].\n- Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6].\n- Step 3: Remove the topmost element = 2. The pile becomes [4,0,6].\n- Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6].\nNote that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves.\n
\n\n

Example 2:

\n\n
\nInput: nums = [2], k = 1\nOutput: -1\nExplanation: \nIn the first move, our only option is to pop the topmost element of the pile.\nSince it is not possible to obtain a non-empty pile after one move, we return -1.\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "K 次操作后最大化顶端元素", "translatedContent": "

给你一个下标从 0 开始的整数数组 nums ，它表示一个 栈 ，其中 nums[0] 是栈顶的元素。

\n\n

每一次操作中，你可以执行以下操作 之一 ：

\n\n\n\n

同时给你一个整数 k ，它表示你总共需要执行操作的次数。

\n\n

请你返回 恰好 执行 k 次操作以后，栈顶元素的 最大值 。如果执行完 k 次操作以后，栈一定为空，请你返回 -1 。

\n\n

 

\n\n

示例 1：

\n\n
\n输入：nums = [5,2,2,4,0,6], k = 4\n输出：5\n解释：\n4 次操作后，栈顶元素为 5 的方法之一为：\n- 第 1 次操作：删除栈顶元素 5 ，栈变为 [2,2,4,0,6] 。\n- 第 2 次操作：删除栈顶元素 2 ，栈变为 [2,4,0,6] 。\n- 第 3 次操作：删除栈顶元素 2 ，栈变为 [4,0,6] 。\n- 第 4 次操作：将 5 添加回栈顶，栈变为 [5,4,0,6] 。\n注意，这不是最后栈顶元素为 5 的唯一方式。但可以证明，4 次操作以后 5 是能得到的最大栈顶元素。\n
\n\n

示例 2：

\n\n
\n输入：nums = [2], k = 1\n输出：-1\n解释：\n第 1 次操作中，我们唯一的选择是将栈顶元素弹出栈。\n由于 1 次操作后无法得到一个非空的栈，所以我们返回 -1 。\n
\n\n

 

\n\n

提示：

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