<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>

import heapq
class Solution:
    def maxEvents(self, events: List[List[int]]) -> int:
        events.sort(key=lambda x: x[0])
        day=1
        index=0
        pq=[]
        ans=0
        total_days= max([event[1] for event in events])
        while day <=total_days:
            # add all events whose start time is before or on the current day
            while index<len(events) and events[index][0]<=day:
                heapq.heappush(pq,events[index][1])
                index+=1 
            # Discard Elements who have ended already 
            while pq and pq[0]<day:
                heapq.heappop(pq)
            # pop the event from the queue which have the most early end time
            if pq:
                heapq.heappop(pq)
                ans+=1
            day+=1
        return ans

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