Two Best Non-Overlapping Events
<p>You are given a <strong>0-indexed</strong> 2D integer array of <code>events</code> where <code>events[i] = [startTime<sub>i</sub>, endTime<sub>i</sub>, value<sub>i</sub>]</code>. The <code>i<sup>th</sup></code> event starts at <code>startTime<sub>i</sub></code><sub> </sub>and ends at <code>endTime<sub>i</sub></code>, and if you attend this event, you will receive a value of <code>value<sub>i</sub></code>. You can choose <strong>at most</strong> <strong>two</strong> <strong>non-overlapping</strong> events to attend such that the sum of their values is <strong>maximized</strong>.</p>

<p>Return <em>this <strong>maximum</strong> sum.</em></p>

<p>Note that the start time and end time is <strong>inclusive</strong>: that is, you cannot attend two events where one of them starts and the other ends at the same time. More specifically, if you attend an event with end time <code>t</code>, the next event must start at or after <code>t + 1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2054.Two%20Best%20Non-Overlapping%20Events/images/picture5.png" style="width: 400px; height: 75px;" />
<pre>
<strong>Input:</strong> events = [[1,3,2],[4,5,2],[2,4,3]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Choose the green events, 0 and 1 for a sum of 2 + 2 = 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="Example 1 Diagram" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2054.Two%20Best%20Non-Overlapping%20Events/images/picture1.png" style="width: 400px; height: 77px;" />
<pre>
<strong>Input:</strong> events = [[1,3,2],[4,5,2],[1,5,5]]
<strong>Output:</strong> 5
<strong>Explanation: </strong>Choose event 2 for a sum of 5.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2054.Two%20Best%20Non-Overlapping%20Events/images/picture3.png" style="width: 400px; height: 66px;" />
<pre>
<strong>Input:</strong> events = [[1,5,3],[1,5,1],[6,6,5]]
<strong>Output:</strong> 8
<strong>Explanation: </strong>Choose events 0 and 2 for a sum of 3 + 5 = 8.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= events.length &lt;= 10<sup>5</sup></code></li>
	<li><code>events[i].length == 3</code></li>
	<li><code>1 &lt;= startTime<sub>i</sub> &lt;= endTime<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= value<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Binary Search; Dynamic Programming; Sorting; Heap (Priority Queue)
func maxTwoEvents(events [][]int) int {
	sort.Slice(events, func(i, j int) bool {
		return events[i][0] < events[j][0]
	})
	n := len(events)
	f := make([]int, n+1)
	for i := n - 1; i >= 0; i-- {
		f[i] = max(f[i+1], events[i][2])
	}
	ans := 0
	for _, e := range events {
		v := e[2]
		left, right := 0, n
		for left < right {
			mid := (left + right) >> 1
			if events[mid][0] > e[1] {
				right = mid
			} else {
				left = mid + 1
			}
		}
		if left < n {
			v += f[left]
		}
		ans = max(ans, v)
	}
	return ans
}