leetcode-problemset/leetcode-cn/originData/maximize-the-minimum-powered-city.json

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            "title": "Maximize the Minimum Powered City",
            "titleSlug": "maximize-the-minimum-powered-city",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>stations</code> of length <code>n</code>, where <code>stations[i]</code> represents the number of power stations in the <code>i<sup>th</sup></code> city.</p>\n\n<p>Each power station can provide power to every city in a fixed <strong>range</strong>. In other words, if the range is denoted by <code>r</code>, then a power station at city <code>i</code> can provide power to all cities <code>j</code> such that <code>|i - j| &lt;= r</code> and <code>0 &lt;= i, j &lt;= n - 1</code>.</p>\n\n<ul>\n\t<li>Note that <code>|x|</code> denotes <strong>absolute</strong> value. For example, <code>|7 - 5| = 2</code> and <code>|3 - 10| = 7</code>.</li>\n</ul>\n\n<p>The <strong>power</strong> of a city is the total number of power stations it is being provided power from.</p>\n\n<p>The government has sanctioned building <code>k</code> more power stations, each of which can be built in any city, and have the same range as the pre-existing ones.</p>\n\n<p>Given the two integers <code>r</code> and <code>k</code>, return <em>the <strong>maximum possible minimum power</strong> of a city, if the additional power stations are built optimally.</em></p>\n\n<p><strong>Note</strong> that you can build the <code>k</code> power stations in multiple cities.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> \nOne of the optimal ways is to install both the power stations at city 1. \nSo stations will become [1,4,4,5,0].\n- City 0 is provided by 1 + 4 = 5 power stations.\n- City 1 is provided by 1 + 4 + 4 = 9 power stations.\n- City 2 is provided by 4 + 4 + 5 = 13 power stations.\n- City 3 is provided by 5 + 4 = 9 power stations.\n- City 4 is provided by 5 + 0 = 5 power stations.\nSo the minimum power of a city is 5.\nSince it is not possible to obtain a larger power, we return 5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> \nIt can be proved that we cannot make the minimum power of a city greater than 4.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == stations.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= stations[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= r&nbsp;&lt;= n - 1</code></li>\n\t<li><code>0 &lt;= k&nbsp;&lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "最大化城市的最小电量",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>stations</code>&nbsp;，其中&nbsp;<code>stations[i]</code>&nbsp;表示第 <code>i</code>&nbsp;座城市的供电站数目。</p>\n\n<p>每个供电站可以在一定 <strong>范围</strong>&nbsp;内给所有城市提供电力。换句话说，如果给定的范围是&nbsp;<code>r</code>&nbsp;，在城市&nbsp;<code>i</code>&nbsp;处的供电站可以给所有满足&nbsp;<code>|i - j| &lt;= r</code> 且&nbsp;<code>0 &lt;= i, j &lt;= n - 1</code>&nbsp;的城市&nbsp;<code>j</code>&nbsp;供电。</p>\n\n<ul>\n\t<li><code>|x|</code>&nbsp;表示 <code>x</code>&nbsp;的 <strong>绝对值</strong>&nbsp;。比方说，<code>|7 - 5| = 2</code>&nbsp;，<code>|3 - 10| = 7</code>&nbsp;。</li>\n</ul>\n\n<p>一座城市的 <strong>电量</strong>&nbsp;是所有能给它供电的供电站数目。</p>\n\n<p>政府批准了可以额外建造 <code>k</code>&nbsp;座供电站，你需要决定这些供电站分别应该建在哪里，这些供电站与已经存在的供电站有相同的供电范围。</p>\n\n<p>给你两个整数&nbsp;<code>r</code> 和&nbsp;<code>k</code>&nbsp;，如果以最优策略建造额外的发电站，返回所有城市中，最小电量的最大值是多少。</p>\n\n<p>这 <code>k</code>&nbsp;座供电站可以建在多个城市。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>stations = [1,2,4,5,0], r = 1, k = 2\n<b>输出：</b>5\n<b>解释：</b>\n最优方案之一是把 2 座供电站都建在城市 1 。\n每座城市的供电站数目分别为 [1,4,4,5,0] 。\n- 城市 0 的供电站数目为 1 + 4 = 5 。\n- 城市 1 的供电站数目为 1 + 4 + 4 = 9 。\n- 城市 2 的供电站数目为 4 + 4 + 5 = 13 。\n- 城市 3 的供电站数目为 5 + 4 = 9 。\n- 城市 4 的供电站数目为 5 + 0 = 5 。\n供电站数目最少是 5 。\n无法得到更优解，所以我们返回 5 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>stations = [4,4,4,4], r = 0, k = 3\n<b>输出：</b>4\n<b>解释：</b>\n无论如何安排，总有一座城市的供电站数目是 4 ，所以最优解是 4 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == stations.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= stations[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= r&nbsp;&lt;= n - 1</code></li>\n\t<li><code>0 &lt;= k&nbsp;&lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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