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Commit f72dc4f

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Sync LeetCode submission Runtime - 173 ms (28.59%), Memory - 53.1 MB (94.66%)
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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>
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<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>
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<p>Return <em>the maximum number of events you can attend</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/02/05/e1.png" style="width: 400px; height: 267px;" />
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<pre>
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<strong>Input:</strong> events = [[1,2],[2,3],[3,4]]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> You can attend all the three events.
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One way to attend them all is as shown.
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Attend the first event on day 1.
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Attend the second event on day 2.
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Attend the third event on day 3.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> events= [[1,2],[2,3],[3,4],[1,2]]
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<strong>Output:</strong> 4
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= events.length &lt;= 10<sup>5</sup></code></li>
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<li><code>events[i].length == 2</code></li>
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<li><code>1 &lt;= startDay<sub>i</sub> &lt;= endDay<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
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</ul>
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# Approach: Greedy
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# n = len(events), T = max(event[1] for event in events)
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# Time: O((T + n) log n)
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# Space: O(n)
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import heapq
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class Solution:
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def maxEvents(self, events: List[List[int]]) -> int:
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n = len(events)
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max_day = max(event[1] for event in events)
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events.sort()
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pq = []
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ans, j = 0, 0
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for i in range(1, max_day + 1):
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while j < n and events[j][0] <= i:
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heapq.heappush(pq, events[j][1])
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j += 1
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while pq and pq[0] < i:
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heapq.heappop(pq)
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if pq:
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heapq.heappop(pq)
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ans += 1
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return ans
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