{ "data": { "question": { "questionId": "3743", "questionFrontendId": "3439", "categoryTitle": "Algorithms", "boundTopicId": 3060967, "title": "Reschedule Meetings for Maximum Free Time I", "titleSlug": "reschedule-meetings-for-maximum-free-time-i", "content": "

You are given an integer eventTime denoting the duration of an event, where the event occurs from time t = 0 to time t = eventTime.

\n\n

You are also given two integer arrays startTime and endTime, each of length n. These represent the start and end time of n non-overlapping meetings, where the ith meeting occurs during the time [startTime[i], endTime[i]].

\n\n

You can reschedule at most k meetings by moving their start time while maintaining the same duration, to maximize the longest continuous period of free time during the event.

\n\n

The relative order of all the meetings should stay the same and they should remain non-overlapping.

\n\n

Return the maximum amount of free time possible after rearranging the meetings.

\n\n

Note that the meetings can not be rescheduled to a time outside the event.

\n\n

 

\n

Example 1:

\n\n
\n

Input: eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5]

\n\n

Output: 2

\n\n

Explanation:

\n\n

\"\"

\n\n

Reschedule the meeting at [1, 2] to [2, 3], leaving no meetings during the time [0, 2].

\n
\n\n

Example 2:

\n\n
\n

Input: eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10]

\n\n

Output: 6

\n\n

Explanation:

\n\n

\"\"

\n\n

Reschedule the meeting at [2, 4] to [1, 3], leaving no meetings during the time [3, 9].

\n
\n\n

Example 3:

\n\n
\n

Input: eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]

\n\n

Output: 0

\n\n

Explanation:

\n\n

There is no time during the event not occupied by meetings.

\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "重新安排会议得到最多空余时间 I", "translatedContent": "

给你一个整数 eventTime 表示一个活动的总时长,这个活动开始于 t = 0 ,结束于 t = eventTime 。

\n\n

同时给你两个长度为 n 的整数数组 startTime 和 endTime 。它们表示这次活动中 n 个时间 没有重叠 的会议,其中第 i 个会议的时间为 [startTime[i], endTime[i]] 。

\n\n

你可以重新安排 至多 k 个会议,安排的规则是将会议时间平移,且保持原来的 会议时长 ,你的目的是移动会议后 最大化 相邻两个会议之间的 最长 连续空余时间。

\n\n

移动前后所有会议之间的 相对 顺序需要保持不变,而且会议时间也需要保持互不重叠。

\n\n

请你返回重新安排会议以后,可以得到的 最大 空余时间。

\n\n

注意,会议 不能 安排到整个活动的时间以外。

\n\n

 

\n\n

示例 1:

\n\n
\n

输入:eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5]

\n\n

输出:2

\n\n

解释:

\n\n

\"\"

\n\n

将 [1, 2] 的会议安排到 [2, 3] ,得到空余时间 [0, 2] 。

\n
\n\n

示例 2:

\n\n
\n

输入:eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10]

\n\n

输出:6

\n\n

解释:

\n\n

\"\"

\n\n

将 [2, 4] 的会议安排到 [1, 3] ,得到空余时间 [3, 9] 。

\n
\n\n

示例 3:

\n\n
\n

输入:eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]

\n\n

输出:0

\n\n

解释:

\n\n

活动中的所有时间都被会议安排满了。

\n
\n\n

 

\n\n

提示:

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Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n

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\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n

\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a-O2 --disable-reflection<\\/code><\\/p>\\r\\n\\r\\n

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