feat: add solution 1751. Maximum Number of Events That Can Be Attended II

Author: tzuyi0817

solutions/1751. Maximum Number of Events That Can Be Attended II/README.md

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+# [1751. Maximum Number of Events That Can Be Attended II](https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended-ii)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given an array of <code>events</code> where <code>events[i] = [startDay<sub>i</sub>, endDay<sub>i</sub>, value<sub>i</sub>]</code>. The <code>i<sup>th</sup></code> event starts at <code>startDay<sub>i</sub></code><sub> </sub>and ends at <code>endDay<sub>i</sub></code>, and if you attend this event, you will receive a value of <code>value<sub>i</sub></code>. You are also given an integer <code>k</code> which represents the maximum number of events you can attend.</p>
+
+<p>You can only attend one event at a time. If you choose to attend an event, you must attend the <strong>entire</strong> event. Note that the end day is <strong>inclusive</strong>: that is, you cannot attend two events where one of them starts and the other ends on the same day.</p>
+
+<p>Return <em>the <strong>maximum sum</strong> of values that you can receive by attending events.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60048-pm.png" style="width: 400px; height: 103px;"></p>
+
+<pre><strong>Input:</strong> events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
+<strong>Output:</strong> 7
+<strong>Explanation: </strong>Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60150-pm.png" style="width: 400px; height: 103px;"></p>
+
+<pre><strong>Input:</strong> events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> Choose event 2 for a total value of 10.
+Notice that you cannot attend any other event as they overlap, and that you do <strong>not</strong> have to attend k events.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60703-pm.png" style="width: 400px; height: 126px;"></strong></p>
+
+<pre><strong>Input:</strong> events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= events.length</code></li>
+	<li><code>1 &lt;= k * events.length &lt;= 10<sup>6</sup></code></li>
+	<li><code>1 &lt;= startDay<sub>i</sub> &lt;= endDay<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= value<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Dynamic Programming + Binary Search`**
+
+- Time complexity: <em>O(nlogn+nk)</em>
+- Space complexity: <em>O(nk)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[][]} events
+ * @param {number} k
+ * @return {number}
+ */
+const maxValue = function (events, k) {
+  const n = events.length;
+  const nextEventMap = new Map();
+  const dp = Array.from({ length: n }, () => new Array(k + 1).fill(-1));
+
+  events.sort((a, b) => a[0] - b[0]);
+
+  const findNextEvent = target => {
+    let left = 0;
+    let right = n - 1;
+
+    while (left <= right) {
+      const mid = Math.floor((left + right) / 2);
+      const start = events[mid][0];
+
+      start > target ? (right = mid - 1) : (left = mid + 1);
+    }
+
+    return left;
+  };
+
+  for (let index = 0; index < n; index++) {
+    const end = events[index][1];
+    const nextEvent = findNextEvent(end);
+
+    nextEventMap.set(index, nextEvent);
+  }
+
+  const attendEvent = (index, quota) => {
+    if (index >= n || quota === 0) return 0;
+    if (dp[index][quota] !== -1) return dp[index][quota];
+    const skipEventValue = attendEvent(index + 1, quota);
+    const value = events[index][2];
+    const nextEvent = nextEventMap.get(index);
+    const nextEventValue = value + attendEvent(nextEvent, quota - 1);
+    const result = Math.max(skipEventValue, nextEventValue);
+
+    dp[index][quota] = result;
+
+    return result;
+  };
+
+  return attendEvent(0, k);
+};
+```

solutions/1751. Maximum Number of Events That Can Be Attended II/solution.js

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+/**
+ * @param {number[][]} events
+ * @param {number} k
+ * @return {number}
+ */
+const maxValue = function (events, k) {
+  const n = events.length;
+  const nextEventMap = new Map();
+  const dp = Array.from({ length: n }, () => new Array(k + 1).fill(-1));
+
+  events.sort((a, b) => a[0] - b[0]);
+
+  const findNextEvent = target => {
+    let left = 0;
+    let right = n - 1;
+
+    while (left <= right) {
+      const mid = Math.floor((left + right) / 2);
+      const start = events[mid][0];
+
+      start > target ? (right = mid - 1) : (left = mid + 1);
+    }
+
+    return left;
+  };
+
+  for (let index = 0; index < n; index++) {
+    const end = events[index][1];
+    const nextEvent = findNextEvent(end);
+
+    nextEventMap.set(index, nextEvent);
+  }
+
+  const attendEvent = (index, quota) => {
+    if (index >= n || quota === 0) return 0;
+    if (dp[index][quota] !== -1) return dp[index][quota];
+    const skipEventValue = attendEvent(index + 1, quota);
+    const value = events[index][2];
+    const nextEvent = nextEventMap.get(index);
+    const nextEventValue = value + attendEvent(nextEvent, quota - 1);
+    const result = Math.max(skipEventValue, nextEventValue);
+
+    dp[index][quota] = result;
+
+    return result;
+  };
+
+  return attendEvent(0, k);
+};