leetcode-problemset/leetcode-cn/originData/find-good-days-to-rob-the-bank.json

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            "questionId": "2205",
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            "categoryTitle": "Algorithms",
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            "title": "Find Good Days to Rob the Bank",
            "titleSlug": "find-good-days-to-rob-the-bank",
            "content": "<p>You and a gang of thieves are planning on robbing a bank. You are given a <strong>0-indexed</strong> integer array <code>security</code>, where <code>security[i]</code> is the number of guards on duty on the <code>i<sup>th</sup></code> day. The days are numbered starting from <code>0</code>. You are also given an integer <code>time</code>.</p>\n\n<p>The <code>i<sup>th</sup></code> day is a good day to rob the bank if:</p>\n\n<ul>\n\t<li>There are at least <code>time</code> days before and after the <code>i<sup>th</sup></code> day,</li>\n\t<li>The number of guards at the bank for the <code>time</code> days <strong>before</strong> <code>i</code> are <strong>non-increasing</strong>, and</li>\n\t<li>The number of guards at the bank for the <code>time</code> days <strong>after</strong> <code>i</code> are <strong>non-decreasing</strong>.</li>\n</ul>\n\n<p>More formally, this means day <code>i</code> is a good day to rob the bank if and only if <code>security[i - time] &gt;= security[i - time + 1] &gt;= ... &gt;= security[i] &lt;= ... &lt;= security[i + time - 1] &lt;= security[i + time]</code>.</p>\n\n<p>Return <em>a list of <strong>all</strong> days <strong>(0-indexed) </strong>that are good days to rob the bank</em>.<em> The order that the days are returned in does<strong> </strong><strong>not</strong> matter.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> security = [5,3,3,3,5,6,2], time = 2\n<strong>Output:</strong> [2,3]\n<strong>Explanation:</strong>\nOn day 2, we have security[0] &gt;= security[1] &gt;= security[2] &lt;= security[3] &lt;= security[4].\nOn day 3, we have security[1] &gt;= security[2] &gt;= security[3] &lt;= security[4] &lt;= security[5].\nNo other days satisfy this condition, so days 2 and 3 are the only good days to rob the bank.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> security = [1,1,1,1,1], time = 0\n<strong>Output:</strong> [0,1,2,3,4]\n<strong>Explanation:</strong>\nSince time equals 0, every day is a good day to rob the bank, so return every day.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> security = [1,2,3,4,5,6], time = 2\n<strong>Output:</strong> []\n<strong>Explanation:</strong>\nNo day has 2 days before it that have a non-increasing number of guards.\nThus, no day is a good day to rob the bank, so return an empty list.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= security.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= security[i], time &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "适合野炊的日子",
            "translatedContent": "<p>你和朋友们准备去野炊。给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>security</code>&nbsp;，其中&nbsp;<code>security[i]</code>&nbsp;是第 <code>i</code>&nbsp;天的建议出行指数。日子从 <code>0</code>&nbsp;开始编号。同时给你一个整数&nbsp;<code>time</code>&nbsp;。</p>\n\n<p>如果第 <code>i</code>&nbsp;天满足以下所有条件，我们称它为一个适合野炊的日子：</p>\n\n<ul>\n\t<li>第 <code>i</code>&nbsp;天前和后都分别至少有 <code>time</code>&nbsp;天。</li>\n\t<li>第 <code>i</code>&nbsp;天前连续 <code>time</code>&nbsp;天建议出行指数都是非递增的。</li>\n\t<li>第 <code>i</code>&nbsp;天后连续 <code>time</code>&nbsp;天建议出行指数都是非递减的。</li>\n</ul>\n\n<p>更正式的，第 <code>i</code> 天是一个适合野炊的日子当且仅当：<code>security[i - time] &gt;= security[i - time + 1] &gt;= ... &gt;= security[i] &lt;= ... &lt;= security[i + time - 1] &lt;= security[i + time]</code>.</p>\n\n<p>请你返回一个数组，包含 <strong>所有</strong> 适合野炊的日子（下标从 <strong>0</strong>&nbsp;开始）。返回的日子可以 <strong>任意</strong>&nbsp;顺序排列。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>security = [5,3,3,3,5,6,2], time = 2\n<b>输出：</b>[2,3]\n<strong>解释：</strong>\n第 2 天，我们有 security[0] &gt;= security[1] &gt;= security[2] &lt;= security[3] &lt;= security[4] 。\n第 3 天，我们有 security[1] &gt;= security[2] &gt;= security[3] &lt;= security[4] &lt;= security[5] 。\n没有其他日子符合这个条件，所以日子 2 和 3 是适合野炊的日子。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>security = [1,1,1,1,1], time = 0\n<b>输出：</b>[0,1,2,3,4]\n<strong>解释：</strong>\n因为 time 等于 0 ，所以每一天都是适合野炊的日子，所以返回每一天。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>security = [1,2,3,4,5,6], time = 2\n<b>输出：</b>[]\n<strong>解释：</strong>\n没有任何一天的前 2 天建议出行指数是非递增的。\n所以没有适合野炊的日子，返回空数组。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= security.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= security[i], time &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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            "hints": [
                "The trivial solution is to check the time days before and after each day. There are a lot of repeated operations using this solution. How could we optimize this solution?",
                "We can use precomputation to make the solution faster.",
                "Use an array to store the number of days before the i<sup>th</sup> day that is non-increasing, and another array to store the number of days after the i<sup>th</sup> day that is non-decreasing."
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