# 1353 Maximum Number of Events That Can Be Attended

You are given an array of `events` where `events[i] = [startDayi, endDayi]`. Every event `i` starts at `startDayi` and ends at `endDayi`.

You can attend an event `i` at any day `d` where `startTimei <= d <= endTimei`. You can only attend one event at any time `d`.

Return the maximum number of events you can attend.

Example 1:

`````` 1 [x _]
2   [x _]
3     [x _]
Day 1 2 3 4

Input: events = [[1,2],[2,3],[3,4]]
Output: 3
Explanation: 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.
``````

Example 2:

``````Input: events= [[1,2],[2,3],[3,4],[1,2]]
Output: 4
``````

Constraints:

• `1 <= events.length <= 105`
• `events[i].length == 2`
• `1 <= startDayi <= endDayi <= 105`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 `````` ``````class Solution: def maxEvents(self, events: List[List[int]]) -> int: n = len(events) events.sort() heap = [] # heap of on-going events' end time res = 0 d = 0 i = 0 # index of sorted events while i < n or heap: # while there are no-yet stated or on-going events if not heap: # no on-going events, jump to the next event d = events[i][0] while i < n and events[i][0] <= d: # add started events' end time to heap heapq.heappush(heap, events[i][1]) i += 1 heapq.heappop(heap) res += 1 d += 1 while heap and heap[0] < d: # remove ended events from heap heapq.heappop(heap) return res``````