leetcode-problemset/leetcode-cn/originData/determine-if-two-events-have-conflict.json

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        "question": {
            "questionId": "2536",
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            "categoryTitle": "Algorithms",
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            "title": "Determine if Two Events Have Conflict",
            "titleSlug": "determine-if-two-events-have-conflict",
            "content": "<p>You are given two arrays of strings that represent two inclusive events that happened <strong>on the same day</strong>, <code>event1</code> and <code>event2</code>, where:</p>\n\n<ul>\n\t<li><code>event1 = [startTime<sub>1</sub>, endTime<sub>1</sub>]</code> and</li>\n\t<li><code>event2 = [startTime<sub>2</sub>, endTime<sub>2</sub>]</code>.</li>\n</ul>\n\n<p>Event times are valid 24 hours format in the form of <code>HH:MM</code>.</p>\n\n<p>A <strong>conflict</strong> happens when two events have some non-empty intersection (i.e., some moment is common to both events).</p>\n\n<p>Return <code>true</code><em> if there is a conflict between two events. Otherwise, return </em><code>false</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> event1 = [&quot;01:15&quot;,&quot;02:00&quot;], event2 = [&quot;02:00&quot;,&quot;03:00&quot;]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> The two events intersect at time 2:00.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> event1 = [&quot;01:00&quot;,&quot;02:00&quot;], event2 = [&quot;01:20&quot;,&quot;03:00&quot;]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> The two events intersect starting from 01:20 to 02:00.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> event1 = [&quot;10:00&quot;,&quot;11:00&quot;], event2 = [&quot;14:00&quot;,&quot;15:00&quot;]\n<strong>Output:</strong> false\n<strong>Explanation:</strong> The two events do not intersect.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>evnet1.length == event2.length == 2.</code></li>\n\t<li><code>event1[i].length == event2[i].length == 5</code></li>\n\t<li><code>startTime<sub>1</sub> &lt;= endTime<sub>1</sub></code></li>\n\t<li><code>startTime<sub>2</sub> &lt;= endTime<sub>2</sub></code></li>\n\t<li>All the event times follow the <code>HH:MM</code> format.</li>\n</ul>\n",
            "translatedTitle": "判断两个事件是否存在冲突",
            "translatedContent": "<p>给你两个字符串数组 <code>event1</code> 和&nbsp;<code>event2</code>&nbsp;，表示发生在同一天的两个闭区间时间段事件，其中：</p>\n\n<ul>\n\t<li><code>event1 = [startTime<sub>1</sub>, endTime<sub>1</sub>]</code> 且</li>\n\t<li><code>event2 = [startTime<sub>2</sub>, endTime<sub>2</sub>]</code></li>\n</ul>\n\n<p>事件的时间为有效的 24 小时制且按&nbsp;<code>HH:MM</code>&nbsp;格式给出。</p>\n\n<p>当两个事件存在某个非空的交集时（即，某些时刻是两个事件都包含的），则认为出现 <strong>冲突</strong>&nbsp;。</p>\n\n<p>如果两个事件之间存在冲突，返回&nbsp;<code>true</code><em>&nbsp;</em>；否则，返回<em>&nbsp;</em><code>false</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><b>示例 1：</b></p>\n\n<pre>\n<b>输入：</b>event1 = [\"01:15\",\"02:00\"], event2 = [\"02:00\",\"03:00\"]\n<b>输出：</b>true\n<b>解释：</b>两个事件在 2:00 出现交集。\n</pre>\n\n<p><b>示例 2：</b></p>\n\n<pre>\n<b>输入：</b>event1 = [\"01:00\",\"02:00\"], event2 = [\"01:20\",\"03:00\"]\n<b>输出：</b>true\n<b>解释：</b>两个事件的交集从 01:20 开始，到 02:00 结束。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>event1 = [\"10:00\",\"11:00\"], event2 = [\"14:00\",\"15:00\"]\n<b>输出：</b>false\n<b>解释：</b>两个事件不存在交集。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><b>提示：</b></p>\n\n<ul>\n\t<li><code>evnet1.length == event2.length == 2.</code></li>\n\t<li><code>event1[i].length == event2[i].length == 5</code></li>\n\t<li><code>startTime<sub>1</sub> &lt;= endTime<sub>1</sub></code></li>\n\t<li><code>startTime<sub>2</sub> &lt;= endTime<sub>2</sub></code></li>\n\t<li>所有事件的时间都按照&nbsp;<code>HH:MM</code>&nbsp;格式给出</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    bool haveConflict(vector<string>& event1, vector<string>& event2) {\n        \n    }\n};",
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                    "code": "-spec have_conflict(Event1 :: [unicode:unicode_binary()], Event2 :: [unicode:unicode_binary()]) -> boolean().\nhave_conflict(Event1, Event2) ->\n  .",
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                "Parse time format to some integer interval first",
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