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# Maximum Number of Events That Can Be Attended

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`. Notice that you can only attend one event at any time `d`.

Return *the maximum number of events* you can attend.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/02/05/e1.png)

```
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
```

**Example 3:**

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

**Example 4:**

```
Input: events = [[1,100000]]
Output: 1
```

**Example 5:**

```
Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
Output: 7
```

**Constraints:**

* `1 <= events.length <= 10^5`
* `events[i].length == 2`
* `1 <= events[i][0] <= events[i][1] <= 10^5`

```java
class Solution {
    public int maxEvents(int[][] events) {
        Arrays.sort(events, (a, b) -> a[0] - b[0]); // sort events increasing by start time
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        int ans = 0, i = 0, n = events.length;
        for (int d = 1; d <= 100000; d++) {
            while (i < n && events[i][0] == d) { // Add new events that can attend on day `d`
                minHeap.add(events[i++][1]);
            }
            while (!minHeap.isEmpty() && minHeap.peek() < d) { // Remove events that are already closed
                minHeap.poll();
            }
            if (!minHeap.isEmpty()) { // Use day `d` to attend to the event that closes earlier
                ans++;
                minHeap.poll();
            }
        }
        return ans;
    }
}
```