mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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175 lines
18 KiB
JSON
175 lines
18 KiB
JSON
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"questionId": "2205",
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"title": "Find Good Days to Rob the Bank",
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"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] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= 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> </p>\n<p><strong>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] >= security[1] >= security[2] <= security[3] <= security[4].\nOn day 3, we have security[1] >= security[2] >= security[3] <= security[4] <= 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>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>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> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= security.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= security[i], time <= 10<sup>5</sup></code></li>\n</ul>\n",
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"code": "# @param {Integer[]} security\n# @param {Integer} time\n# @return {Integer[]}\ndef good_days_to_rob_bank(security, time)\n \nend",
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"code": "impl Solution {\n pub fn good_days_to_rob_bank(security: Vec<i32>, time: i32) -> Vec<i32> {\n \n }\n}",
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"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?",
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"We can use precomputation to make the solution faster.",
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"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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