leetcode-problemset/leetcode-cn/originData/maximize-the-topmost-element-after-k-moves.json

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            "title": "Maximize the Topmost Element After K Moves",
            "titleSlug": "maximize-the-topmost-element-after-k-moves",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> representing the contents of a <b>pile</b>, where <code>nums[0]</code> is the topmost element of the pile.</p>\n\n<p>In one move, you can perform <strong>either</strong> of the following:</p>\n\n<ul>\n\t<li>If the pile is not empty, <strong>remove</strong> the topmost element of the pile.</li>\n\t<li>If there are one or more removed elements, <strong>add</strong> any one of them back onto the pile. This element becomes the new topmost element.</li>\n</ul>\n\n<p>You are also given an integer <code>k</code>, which denotes the total number of moves to be made.</p>\n\n<p>Return <em>the <strong>maximum value</strong> of the topmost element of the pile possible after <strong>exactly</strong></em> <code>k</code> <em>moves</em>. In case it is not possible to obtain a non-empty pile after <code>k</code> moves, return <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,2,2,4,0,6], k = 4\n<strong>Output:</strong> 5\n<strong>Explanation:</strong>\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</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2], k = 1\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> \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</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i], k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "K 次操作后最大化顶端元素",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，它表示一个 <strong>栈</strong> ，其中 <code>nums[0]</code>&nbsp;是栈顶的元素。</p>\n\n<p>每一次操作中，你可以执行以下操作 <strong>之一</strong>&nbsp;：</p>\n\n<ul>\n\t<li>如果栈非空，那么 <strong>删除</strong>&nbsp;栈顶端的元素。</li>\n\t<li>如果存在 1 个或者多个被删除的元素，你可以从它们中选择任何一个，<b>添加</b>&nbsp;回栈顶，这个元素成为新的栈顶元素。</li>\n</ul>\n\n<p>同时给你一个整数&nbsp;<code>k</code>&nbsp;，它表示你总共需要执行操作的次数。</p>\n\n<p>请你返回 <strong>恰好</strong>&nbsp;执行 <code>k</code>&nbsp;次操作以后，栈顶元素的 <strong>最大值</strong>&nbsp;。如果执行完 <code>k</code>&nbsp;次操作以后，栈一定为空，请你返回 <code>-1</code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [5,2,2,4,0,6], k = 4\n<b>输出：</b>5\n<strong>解释：</strong>\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</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [2], k = 1\n<b>输出：</b>-1\n<b>解释：</b>\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</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i], k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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