{ "data": { "question": { "questionId": "2205", "questionFrontendId": "2100", "categoryTitle": "Algorithms", "boundTopicId": 1149825, "title": "Find Good Days to Rob the Bank", "titleSlug": "find-good-days-to-rob-the-bank", "content": "

You and a gang of thieves are planning on robbing a bank. You are given a 0-indexed integer array security, where security[i] is the number of guards on duty on the ith day. The days are numbered starting from 0. You are also given an integer time.

\n\n

The ith day is a good day to rob the bank if:

\n\n\n\n

More formally, this means day i is a good day to rob the bank if and only if security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

\n\n

Return a list of all days (0-indexed) that are good days to rob the bank. The order that the days are returned in does not matter.

\n\n

 

\n

Example 1:

\n\n
\nInput: security = [5,3,3,3,5,6,2], time = 2\nOutput: [2,3]\nExplanation:\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
\n\n

Example 2:

\n\n
\nInput: security = [1,1,1,1,1], time = 0\nOutput: [0,1,2,3,4]\nExplanation:\nSince time equals 0, every day is a good day to rob the bank, so return every day.\n
\n\n

Example 3:

\n\n
\nInput: security = [1,2,3,4,5,6], time = 2\nOutput: []\nExplanation:\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
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "适合打劫银行的日子", "translatedContent": "

你和一群强盗准备打劫银行。给你一个下标从 0 开始的整数数组 security ,其中 security[i] 是第 i 天执勤警卫的数量。日子从 0 开始编号。同时给你一个整数 time 。

\n\n

如果第 i 天满足以下所有条件,我们称它为一个适合打劫银行的日子:

\n\n\n\n

更正式的,第 i 天是一个合适打劫银行的日子当且仅当:security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

\n\n

请你返回一个数组,包含 所有 适合打劫银行的日子(下标从 0 开始)。返回的日子可以 任意 顺序排列。

\n\n

 

\n\n

示例 1:

\n\n
\n输入:security = [5,3,3,3,5,6,2], time = 2\n输出:[2,3]\n解释:\n第 2 天,我们有 security[0] >= security[1] >= security[2] <= security[3] <= security[4] 。\n第 3 天,我们有 security[1] >= security[2] >= security[3] <= security[4] <= security[5] 。\n没有其他日子符合这个条件,所以日子 2 和 3 是适合打劫银行的日子。\n
\n\n

示例 2:

\n\n
\n输入:security = [1,1,1,1,1], time = 0\n输出:[0,1,2,3,4]\n解释:\n因为 time 等于 0 ,所以每一天都是适合打劫银行的日子,所以返回每一天。\n
\n\n

示例 3:

\n\n
\n输入:security = [1,2,3,4,5,6], time = 2\n输出:[]\n解释:\n没有任何一天的前 2 天警卫数目是非递增的。\n所以没有适合打劫银行的日子,返回空数组。\n
\n\n

 

\n\n

提示:

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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 ith day that is non-increasing, and another array to store the number of days after the ith day that is non-decreasing." ], "solution": null, "status": null, "sampleTestCase": "[5,3,3,3,5,6,2]\n2", "metaData": "{\n \"name\": \"goodDaysToRobBank\",\n \"params\": [\n {\n \"type\": \"integer[]\",\n \"name\": \"security\"\n },\n {\n \"type\": \"integer\",\n \"name\": \"time\"\n }\n ],\n \"return\": {\n \"type\": \"list\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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