How do you solve Two Best Non-Overlapping Events on LeetCode?

Two Best Non-Overlapping Events (LeetCode #2054) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting events by end time allows us to efficiently find non-overlapping events.

What is the brute force solution for Two Best Non-Overlapping Events?

The brute force approach for Two Best Non-Overlapping Events is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of events to see if they overlap and calculating their total value. This is straightforward but inefficient, especially for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Two Best Non-Overlapping Events?

Always consider sorting as a first step when dealing with intervals. Explain your thought process clearly; it helps interviewers understand your approach. Practice identifying overlapping intervals in different contexts.

What are common mistakes when solving Two Best Non-Overlapping Events?

Failing to account for inclusive start and end times when checking for overlaps. Not using sorting effectively to simplify the problem.

#2054

Two Best Non-Overlapping Events

Medium

ArrayBinary SearchDynamic ProgrammingSortingHeap (Priority Queue)Dynamic ProgrammingBinary SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting the events by their end times and using a dynamic programming approach, we can efficiently find the best non-overlapping events. This allows us to quickly determine the maximum value of events that do not overlap with the current event.

⚙️

Algorithm

3 steps

1Step 1: Sort the events based on their end times.
2Step 2: Use a dynamic programming array to store the maximum value obtainable up to each event.
3Step 3: For each event, find the last non-overlapping event using binary search and update the DP array accordingly.

solution.py16 lines

1# Full working Python code
2import bisect
3
4def maxTwoEvents(events):
5    events.sort(key=lambda x: x[1])
6    n = len(events)
7    dp = [0] * n
8    for i in range(n):
9        dp[i] = events[i][2]
10        if i > 0:
11            dp[i] = max(dp[i], dp[i-1])
12        # Find the last non-overlapping event
13        j = bisect.bisect_right(events, [events[i][0] - 1]) - 1
14        if j >= 0:
15            dp[i] = max(dp[i], events[i][2] + (dp[j] if j >= 0 else 0))
16    return max(dp)

ℹ

Complexity note: The sorting step takes O(n log n), and filling the DP array takes O(n), leading to an overall time complexity of O(n log n). The space complexity is O(n) due to the DP array.

1Sorting events by end time allows us to efficiently find non-overlapping events.
2Using dynamic programming helps to build solutions incrementally, maximizing values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.