leetcode-problemset/leetcode-cn/originData/finding-the-users-active-minutes.json

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        "question": {
            "questionId": "1945",
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            "title": "Finding the Users Active Minutes",
            "titleSlug": "finding-the-users-active-minutes",
            "content": "<p>You are given the logs for users&#39; actions on LeetCode, and an integer <code>k</code>. The logs are represented by a 2D integer array <code>logs</code> where each <code>logs[i] = [ID<sub>i</sub>, time<sub>i</sub>]</code> indicates that the user with <code>ID<sub>i</sub></code> performed an action at the minute <code>time<sub>i</sub></code>.</p>\n\n<p><strong>Multiple users</strong> can perform actions simultaneously, and a single user can perform <strong>multiple actions</strong> in the same minute.</p>\n\n<p>The <strong>user active minutes (UAM)</strong> for a given user is defined as the <strong>number of unique minutes</strong> in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.</p>\n\n<p>You are to calculate a <strong>1-indexed</strong> array <code>answer</code> of size <code>k</code> such that, for each <code>j</code> (<code>1 &lt;= j &lt;= k</code>), <code>answer[j]</code> is the <strong>number of users</strong> whose <strong>UAM</strong> equals <code>j</code>.</p>\n\n<p>Return <i>the array </i><code>answer</code><i> as described above</i>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5\n<strong>Output:</strong> [0,2,0,0,0]\n<strong>Explanation:</strong>\nThe user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).\nThe user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.\nSince both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> logs = [[1,1],[2,2],[2,3]], k = 4\n<strong>Output:</strong> [1,1,0,0]\n<strong>Explanation:</strong>\nThe user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.\nThe user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.\nThere is one user with a UAM of 1 and one with a UAM of 2.\nHence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= logs.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= ID<sub>i</sub> &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>k</code> is in the range <code>[The maximum <strong>UAM</strong> for a user, 10<sup>5</sup>]</code>.</li>\n</ul>\n",
            "translatedTitle": "查找用户活跃分钟数",
            "translatedContent": "<p>给你用户在 LeetCode 的操作日志，和一个整数 <code>k</code> 。日志用一个二维整数数组 <code>logs</code> 表示，其中每个 <code>logs[i] = [ID<sub>i</sub>, time<sub>i</sub>]</code> 表示 ID 为 <code>ID<sub>i</sub></code> 的用户在 <code>time<sub>i</sub></code> 分钟时执行了某个操作。</p>\n\n<p><strong>多个用户 </strong>可以同时执行操作，单个用户可以在同一分钟内执行 <strong>多个操作</strong> 。</p>\n\n<p>指定用户的<strong> 用户活跃分钟数（user active minutes，UAM）</strong> 定义为用户对 LeetCode 执行操作的 <strong>唯一分钟数</strong> 。 即使一分钟内执行多个操作，也只能按一分钟计数。</p>\n\n<p>请你统计用户活跃分钟数的分布情况，统计结果是一个长度为 <code>k</code> 且 <strong>下标从 1 开始计数</strong> 的数组 <code>answer</code> ，对于每个 <code>j</code>（<code>1 <= j <= k</code>），<code>answer[j]</code> 表示 <strong>用户活跃分钟数</strong> 等于 <code>j</code> 的用户数。</p>\n\n<p>返回上面描述的答案数组<i> </i><code>answer</code><i> </i>。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5\n<strong>输出：</strong>[0,2,0,0,0]\n<strong>解释：</strong>\nID=0 的用户执行操作的分钟分别是：5 、2 和 5 。因此，该用户的用户活跃分钟数为 2（分钟 5 只计数一次）\nID=1 的用户执行操作的分钟分别是：2 和 3 。因此，该用户的用户活跃分钟数为 2\n2 个用户的用户活跃分钟数都是 2 ，answer[2] 为 2 ，其余 answer[j] 的值都是 0\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>logs = [[1,1],[2,2],[2,3]], k = 4\n<strong>输出：</strong>[1,1,0,0]\n<strong>解释：</strong>\nID=1 的用户仅在分钟 1 执行单个操作。因此，该用户的用户活跃分钟数为 1\nID=2 的用户执行操作的分钟分别是：2 和 3 。因此，该用户的用户活跃分钟数为 2\n1 个用户的用户活跃分钟数是 1 ，1 个用户的用户活跃分钟数是 2 \n因此，answer[1] = 1 ，answer[2] = 1 ，其余的值都是 0\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= logs.length <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= ID<sub>i</sub> <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= time<sub>i</sub> <= 10<sup>5</sup></code></li>\n\t<li><code>k</code> 的取值范围是 <code>[用户的最大用户活跃分钟数, 10<sup>5</sup>]</code></li>\n</ul>\n",
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