Leetcode 2054 two best non overlapping events

You are given a 0-indexed 2D integer array of events where events[i] = [startTimei, endTimei, valuei]. The ith event starts at startTimei and ends at endTimei, and if you attend this event, you will receive a value of valuei. You can choose at most two non-overlapping events to attend such that the sum of their values is maximized.

Return this maximum sum.

Note that the start time and end time is inclusive: 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 t, the next event must start at or after t + 1.

Example 1:

Input: events = [[1,3,2],[4,5,2],[2,4,3]]
Output: 4
Explanation: Choose the green events, 0 and 1 for a sum of 2 + 2 = 4.

Example 2:

Input: events = [[1,3,2],[4,5,2],[1,5,5]]
Output: 5
Explanation: Choose event 2 for a sum of 5.

Example 3:

Input: events = [[1,5,3],[1,5,1],[6,6,5]]
Output: 8
Explanation: Choose events 0 and 2 for a sum of 3 + 5 = 8

Constraints:

• 2 <= events.length <= 105
• events[i].length == 3
• 1 <= startTimei <= endTimei <= 109
• 1 <= valuei <= 106

Topic link ：https://leetcode.com/problems/two-best-non-overlapping-events/

The main idea of the topic ： At most two disjoint intervals can be selected , Ask for it value The maximum value of and

Topic analysis ： Sort left endpoint , Preprocess each section to the rightmost maximum value value , Enumerate left endpoints , Split the right end point

40ms, Time beats 81.8%

class Solution {
    
    class Event implements Comparable {
        int st, ed, val;
        Event(int st, int ed, int val) {
            this.st = st;
            this.ed = ed;
            this.val = val;
        }
        
        @Override
        public int compareTo(Object o) {
            Event e = (Event) o;
            if (this.st > e.st) {
                return 1;
            } else if (this.st < e.st) {
                return -1;
            }
            return 0;
        }
    }
    
    // find pos of first event which st value larger than x
    int bsearch(Event[] e, int l, int r, int x) {
        int mid = 0, ans = -1;
        while (l <= r) {
            mid = (l + r) >> 1;
            if (e[mid].st > x) {
                ans = mid;
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return ans;
    }
    
    public int maxTwoEvents(int[][] events) {
        int n = events.length;
        Event[] e = new Event[n];
        for (int i = 0; i < n; i++) {
            e[i] = new Event(events[i][0], events[i][1], events[i][2]);
        }
        Arrays.sort(e);
        int cur = 0;
        int[] ma = new int[n];
        for (int i = n - 1; i >= 0; i--) {
            if (cur < e[i].val) {
                cur = e[i].val;
            }
            ma[i] = cur;
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int pos = bsearch(e, i, n - 1, e[i].ed);
            if (pos != -1) {
                ans = Math.max(ans, e[i].val + ma[pos]);
            }
        }
        return Math.max(ma[0], ans);
    }
}