2054. Two Best Non-Overlapping Events
Description
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
Solution
two-best-non-overlapping-events.py
class Solution:
def maxTwoEvents(self, events: List[List[int]]) -> int:
A = []
for start, end, value in events:
A.append((start, 1, value))
A.append((end + 1, -1, value))
A.sort()
currMax = res = 0
for curr, t, x in A:
if t == 1:
res = max(res, currMax + x)
else:
currMax = max(currMax, x)
return res