leetcode-problemset/leetcode-cn/originData/maximum-points-after-enemy-battles.json

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        "question": {
            "questionId": "3264",
            "questionFrontendId": "3207",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Points After Enemy Battles",
            "titleSlug": "maximum-points-after-enemy-battles",
            "content": "<p>You are given an integer array <code>enemyEnergies</code> denoting the energy values of various enemies.</p>\n\n<p>You are also given an integer <code>currentEnergy</code> denoting the amount of energy you have initially.</p>\n\n<p>You start with 0 points, and all the enemies are unmarked initially.</p>\n\n<p>You can perform <strong>either</strong> of the following operations <strong>zero </strong>or multiple times to gain points:</p>\n\n<ul>\n\t<li>Choose an <strong>unmarked</strong> enemy, <code>i</code>, such that <code>currentEnergy &gt;= enemyEnergies[i]</code>. By choosing this option:\n\n\t<ul>\n\t\t<li>You gain 1 point.</li>\n\t\t<li>Your energy is reduced by the enemy&#39;s energy, i.e. <code>currentEnergy = currentEnergy - enemyEnergies[i]</code>.</li>\n\t</ul>\n\t</li>\n\t<li>If you have <strong>at least</strong> 1 point, you can choose an <strong>unmarked</strong> enemy, <code>i</code>. By choosing this option:\n\t<ul>\n\t\t<li>Your energy increases by the enemy&#39;s energy, i.e. <code>currentEnergy = currentEnergy + enemyEnergies[i]</code>.</li>\n\t\t<li>The <font face=\"monospace\">e</font>nemy <code>i</code> is <strong>marked</strong>.</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>Return an integer denoting the <strong>maximum</strong> points you can get in the end by optimally performing operations.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">enemyEnergies = [3,2,2], currentEnergy = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">3</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The following operations can be performed to get 3 points, which is the maximum:</p>\n\n<ul>\n\t<li>First operation on enemy 1: <code>points</code> increases by 1, and <code>currentEnergy</code> decreases by 2. So, <code>points = 1</code>, and <code>currentEnergy = 0</code>.</li>\n\t<li>Second operation on enemy 0: <code>currentEnergy</code> increases by 3, and enemy 0 is marked. So, <code>points = 1</code>, <code>currentEnergy = 3</code>, and marked enemies = <code>[0]</code>.</li>\n\t<li>First operation on enemy 2: <code>points</code> increases by 1, and <code>currentEnergy</code> decreases by 2. So, <code>points = 2</code>, <code>currentEnergy = 1</code>, and marked enemies = <code>[0]</code>.</li>\n\t<li>Second operation on enemy 2: <code>currentEnergy</code> increases by 2, and enemy 2 is marked. So, <code>points = 2</code>, <code>currentEnergy = 3</code>, and marked enemies = <code>[0, 2]</code>.</li>\n\t<li>First operation on enemy 1: <code>points</code> increases by 1, and <code>currentEnergy</code> decreases by 2. So, <code>points = 3</code>, <code>currentEnergy = 1</code>, and marked enemies = <code>[0, 2]</code>.</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">enemyEnergies = </span>[2]<span class=\"example-io\">, currentEnergy = 10</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>Explanation: </strong></p>\n\n<p>Performing the first operation 5 times on enemy 0 results in the maximum number of points.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= enemyEnergies.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= enemyEnergies[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= currentEnergy &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "与敌人战斗后的最大分数",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>enemyEnergies</code>&nbsp;，它表示一个下标从 <strong>0</strong>&nbsp;开始的敌人能量数组。</p>\n\n<p>同时给你一个整数&nbsp;<code>currentEnergy</code>&nbsp;，它表示你一开始拥有的能量值总量。</p>\n\n<p>你一开始的分数为&nbsp;<code>0</code>&nbsp;，且一开始所有的敌人都未标记。</p>\n\n<p>你可以通过以下操作 <b>之一</b>&nbsp;<strong>任意次</strong>（也可以&nbsp;<strong>0</strong>&nbsp;次）来得分：</p>\n\n<ul>\n\t<li>选择一个 <strong>未标记</strong>&nbsp;且满足&nbsp;<code>currentEnergy &gt;= enemyEnergies[i]</code>&nbsp;的敌人&nbsp;<code>i</code>&nbsp;。在这个操作中：\n\n\t<ul>\n\t\t<li>你会获得 <code>1</code>&nbsp;分。</li>\n\t\t<li>你的能量值减少&nbsp;<code>enemyEnergies[i]</code>&nbsp;，也就是说&nbsp;<code>currentEnergy = currentEnergy - enemyEnergies[i]</code>&nbsp;。</li>\n\t</ul>\n\t</li>\n\t<li>如果你目前&nbsp;<strong>至少</strong>&nbsp;有 <code>1</code>&nbsp;分，你可以选择一个&nbsp;<strong>未标记</strong>&nbsp;的敌人&nbsp;<code>i</code>&nbsp;。在这个操作中：\n\t<ul>\n\t\t<li>你的能量值增加 <code>enemyEnergies[i]</code>&nbsp;，也就是说&nbsp;<code>currentEnergy = currentEnergy + enemyEnergies[i]</code>&nbsp;。</li>\n\t\t<li>敌人&nbsp;<code>i</code> <strong>被标记</strong>&nbsp;。</li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>请你返回通过以上操作，<strong>最多</strong>&nbsp;可以获得多少分。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><b>输入：</b>enemyEnergies = [3,2,2], currentEnergy = 2</p>\n\n<p><b>输出：</b>3</p>\n\n<p><strong>解释：</strong></p>\n\n<p>通过以下操作可以得到最大得分 3 分：</p>\n\n<ul>\n\t<li>对敌人 1 使用第一种操作：<code>points</code>&nbsp;增加 1 ，<code>currentEnergy</code>&nbsp;减少 2 。所以&nbsp;<code>points = 1</code>&nbsp;且&nbsp;<code>currentEnergy = 0</code>&nbsp;。</li>\n\t<li>对敌人 0 使用第二种操作：<code>currentEnergy</code>&nbsp;增加 3 ，敌人 0 被标记。所以&nbsp;<code>points = 1</code>&nbsp;，<code>currentEnergy = 3</code>&nbsp;，被标记的敌人包括&nbsp;<code>[0]</code>&nbsp;。</li>\n\t<li>对敌人 2 使用第一种操作：<code>points</code>&nbsp;增加 1 ，<code>currentEnergy</code>&nbsp;减少 2 。所以&nbsp;<code>points = 2</code>&nbsp;且&nbsp;<code>currentEnergy = 1</code>&nbsp;，被标记的敌人包括<code>[0]</code>&nbsp;。</li>\n\t<li>对敌人 2 使用第二种操作：<code>currentEnergy</code>&nbsp;增加 2 ，敌人 2 被标记。所以&nbsp;<code>points = 2</code>&nbsp;，<code>currentEnergy = 3</code>&nbsp;且被标记的敌人包括&nbsp;<code>[0, 2]</code>&nbsp;。</li>\n\t<li>对敌人 1 使用第一种操作：<code>points</code>&nbsp;增加 1 ，<code>currentEnergy</code>&nbsp;减少 2 。所以&nbsp;<code>points = 3</code>&nbsp;，<code>currentEnergy = 1</code>&nbsp;，被标记的敌人包括&nbsp;<code>[0, 2]</code>&nbsp;。</li>\n</ul>\n\n<p><strong>示例 2：</strong></p>\n\n<p><b>输入：</b>enemyEnergies =&nbsp;[2], currentEnergy = 10</p>\n\n<p><b>输出：</b>5</p>\n\n<p><strong>解释：</strong></p>\n\n<p>通过对敌人 0 进行第一种操作 5 次，得到最大得分。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= enemyEnergies.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= enemyEnergies[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= currentEnergy &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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