leetcode-problemset/leetcode-cn/originData/reschedule-meetings-for-maximum-free-time-i.json

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        "question": {
            "questionId": "3743",
            "questionFrontendId": "3439",
            "categoryTitle": "Algorithms",
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            "title": "Reschedule Meetings for Maximum Free Time I",
            "titleSlug": "reschedule-meetings-for-maximum-free-time-i",
            "content": "<p>You are given an integer <code>eventTime</code> denoting the duration of an event, where the event occurs from time <code>t = 0</code> to time <code>t = eventTime</code>.</p>\n\n<p>You are also given two integer arrays <code>startTime</code> and <code>endTime</code>, each of length <code>n</code>. These represent the start and end time of <code>n</code> <strong>non-overlapping</strong> meetings, where the <code>i<sup>th</sup></code> meeting occurs during the time <code>[startTime[i], endTime[i]]</code>.</p>\n\n<p>You can reschedule <strong>at most</strong> <code>k</code> meetings by moving their start time while maintaining the <strong>same duration</strong>, to <strong>maximize</strong> the <strong>longest</strong> <em>continuous period of free time</em> during the event.</p>\n\n<p>The <strong>relative</strong> order of all the meetings should stay the<em> same</em> and they should remain non-overlapping.</p>\n\n<p>Return the <strong>maximum</strong> amount of free time possible after rearranging the meetings.</p>\n\n<p><strong>Note</strong> that the meetings can <strong>not</strong> be rescheduled to a time outside the event.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/12/21/example0_rescheduled.png\" style=\"width: 375px; height: 123px;\" /></p>\n\n<p>Reschedule the meeting at <code>[1, 2]</code> to <code>[2, 3]</code>, leaving no meetings during the time <code>[0, 2]</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">6</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/12/21/example1_rescheduled.png\" style=\"width: 375px; height: 125px;\" /></p>\n\n<p>Reschedule the meeting at <code>[2, 4]</code> to <code>[1, 3]</code>, leaving no meetings during the time <code>[3, 9]</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There is no time during the event not occupied by meetings.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= eventTime &lt;= 10<sup>9</sup></code></li>\n\t<li><code>n == startTime.length == endTime.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n</code></li>\n\t<li><code>0 &lt;= startTime[i] &lt; endTime[i] &lt;= eventTime</code></li>\n\t<li><code>endTime[i] &lt;= startTime[i + 1]</code> where <code>i</code> lies in the range <code>[0, n - 2]</code>.</li>\n</ul>\n",
            "translatedTitle": "重新安排会议得到最多空余时间 I",
            "translatedContent": "<p>给你一个整数&nbsp;<code>eventTime</code>&nbsp;表示一个活动的总时长，这个活动开始于&nbsp;<code>t = 0</code>&nbsp;，结束于&nbsp;<code>t = eventTime</code>&nbsp;。</p>\n\n<p>同时给你两个长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>startTime</code> 和&nbsp;<code>endTime</code>&nbsp;。它们表示这次活动中 <code>n</code>&nbsp;个时间&nbsp;<strong>没有重叠</strong>&nbsp;的会议，其中第&nbsp;<code>i</code>&nbsp;个会议的时间为&nbsp;<code>[startTime[i], endTime[i]]</code>&nbsp;。</p>\n\n<p>你可以重新安排 <strong>至多</strong>&nbsp;<code>k</code>&nbsp;个会议，安排的规则是将会议时间平移，且保持原来的 <strong>会议时长</strong>&nbsp;，你的目的是移动会议后 <strong>最大化</strong>&nbsp;相邻两个会议之间的 <strong>最长</strong> 连续空余时间。</p>\n\n<p>移动前后所有会议之间的 <strong>相对</strong>&nbsp;顺序需要保持不变，而且会议时间也需要保持互不重叠。</p>\n\n<p>请你返回重新安排会议以后，可以得到的 <strong>最大</strong>&nbsp;空余时间。</p>\n\n<p><b>注意</b>，会议 <strong>不能</strong>&nbsp;安排到整个活动的时间以外。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/12/21/example0_rescheduled.png\" style=\"width: 375px; height: 123px;\" /></p>\n\n<p>将&nbsp;<code>[1, 2]</code>&nbsp;的会议安排到&nbsp;<code>[2, 3]</code>&nbsp;，得到空余时间&nbsp;<code>[0, 2]</code>&nbsp;。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>6</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/12/21/example1_rescheduled.png\" style=\"width: 375px; height: 125px;\" /></p>\n\n<p>将&nbsp;<code>[2, 4]</code>&nbsp;的会议安排到&nbsp;<code>[1, 3]</code>&nbsp;，得到空余时间&nbsp;<code>[3, 9]</code>&nbsp;。</p>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>0</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>活动中的所有时间都被会议安排满了。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= eventTime &lt;= 10<sup>9</sup></code></li>\n\t<li><code>n == startTime.length == endTime.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n</code></li>\n\t<li><code>0 &lt;= startTime[i] &lt; endTime[i] &lt;= eventTime</code></li>\n\t<li><code>endTime[i] &lt;= startTime[i + 1]</code> 其中&nbsp;<code>i</code>&nbsp;在范围&nbsp;<code>[0, n - 2]</code>&nbsp;之间。</li>\n</ul>\n",
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                "In a sequence of <code>K</code> meetings and <code>K + 1</code> gaps, you could move all meetings to the start of the sequence to get the max free time.",
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