Sync LeetCode submission Runtime - 173 ms (28.59%), Memory - 53.1 MB (94.66%)

Author: jimit105

1478-maximum-number-of-events-that-can-be-attended/README.md

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+<p>You are given an array of <code>events</code> where <code>events[i] = [startDay<sub>i</sub>, endDay<sub>i</sub>]</code>. Every event <code>i</code> starts at <code>startDay<sub>i</sub></code><sub> </sub>and ends at <code>endDay<sub>i</sub></code>.</p>
+
+<p>You can attend an event <code>i</code> at any day <code>d</code> where <code>startTime<sub>i</sub> &lt;= d &lt;= endTime<sub>i</sub></code>. You can only attend one event at any time <code>d</code>.</p>
+
+<p>Return <em>the maximum number of events you can attend</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/02/05/e1.png" style="width: 400px; height: 267px;" />
+<pre>
+<strong>Input:</strong> events = [[1,2],[2,3],[3,4]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> You can attend all the three events.
+One way to attend them all is as shown.
+Attend the first event on day 1.
+Attend the second event on day 2.
+Attend the third event on day 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> events= [[1,2],[2,3],[3,4],[1,2]]
+<strong>Output:</strong> 4
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= events.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>events[i].length == 2</code></li>
+	<li><code>1 &lt;= startDay<sub>i</sub> &lt;= endDay<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
+</ul>

1478-maximum-number-of-events-that-can-be-attended/solution.py

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+# Approach: Greedy
+
+# n = len(events), T = max(event[1] for event in events)
+# Time: O((T + n) log n)
+# Space: O(n)
+
+import heapq
+
+class Solution:
+    def maxEvents(self, events: List[List[int]]) -> int:
+        n = len(events)
+        max_day = max(event[1] for event in events)
+        events.sort()
+        pq = []
+        ans, j = 0, 0
+
+        for i in range(1, max_day + 1):
+            while j < n and events[j][0] <= i:
+                heapq.heappush(pq, events[j][1])
+                j += 1
+            while pq and pq[0] < i:
+                heapq.heappop(pq)
+            if pq:
+                heapq.heappop(pq)
+                ans += 1
+
+        return ans
+