When juggling multiple opportunities, the trick lies in choosing the ones that offer the highest rewards while avoiding overlaps. This problem is a classic example of finding an optimal solution under constraints, where we aim to maximize the sum of values by attending at most two non-overlapping events. 💡 Let’s unravel it step by step and dive into the magic of efficient algorithms! 🚀

💼 Problem Statement

You are given a 0-indexed 2D integer array events where:

• events[i] = [startTimeᵢ, endTimeᵢ, valueᵢ],
• startTimeᵢ is the start time of the i-th event,
• endTimeᵢ is the end time of the i-th event,
• valueᵢ is the value obtained from attending the i-th event.

Goal 🎯

Select at most two non-overlapping events such that the sum of their values is maximized.

Key Constraints:

1. Start and end times are inclusive, meaning you cannot attend two events where one starts as the other ends. Specifically:
   • If an event ends at time t, the next event must start at or after t + 1.
2. Return the maximum sum of values from two non-overlapping events.

✨ Example 1:

Input:

events = [[1,3,2],[4,5,2],[2,4,3]]

Output:

4

Explanation:

Choose events [1,3,2] and [4,5,2] for a total value of 2 + 2 = 4.

✨ Example 2:

Input:

events = [[1,3,2],[4,5,2],[1,5,5]]

Output:

5

Explanation:

Choose the single event [1,5,5] since attending any other event would cause overlap.

✨ Example 3:

Input:

events = [[1,5,3],[1,5,1],[6,6,5]]

Output:

8

Explanation:

Choose events [1,5,3] and [6,6,5] for a total value of 3 + 5 = 8.

🧩 Edge Cases to Consider

1. Small Inputs
   • Only two events that overlap:

     events = [[1,2,3],[2,3,5]]
     Output: 5
     

   • Only one event:

     events = [[1,5,10]]
     Output: 10
     

2. All Events Overlap
   • Example:

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

     In this case, you can pick only one event with the highest value.

3. Disjoint Events
   • Example:

     events = [[1,2,10],[3,4,20],[5,6,30]]
     Output: 50
     

     Here, you can pick the two highest-value events as they don’t overlap.

4. Sparse Time Range
   • Events that are widely spaced apart:

     events = [[1,2,1],[1000,1001,2],[2000,2001,3]]
     Output: 5
     

🔍 Key Insights

1. Sorting for Simplicity
   Sorting events by their end times ensures that we process events in a timeline order. This makes it easier to compare overlapping conditions.

2. Binary Search for Efficiency
   When checking for the highest value of a previously attended event that doesn’t overlap with the current event, a binary search helps find the best match quickly.

3. Dynamic Programming for Optimization
   By maintaining a rolling record of maximum event values, we avoid recomputation and make the solution efficient.

🛠️ Optimized Solution

📚 Detailed Solution Approach

The goal is to maximize the sum of values from at most two non-overlapping events. The key challenges here are:

1. Efficiently checking which events can be combined without overlapping.
2. Tracking the highest values of possible event combinations as we iterate through the list.

Let’s break down the solution step by step.

🔑 Step 1: Sort Events by End Time

Sorting events by their end times simplifies the problem. After sorting:

• Any previously processed event is guaranteed to end before the current event.
• This ensures that we only check whether the current event overlaps with earlier events and not vice versa.

For example, if the input events are:

events = [[1, 3, 5], [2, 5, 7], [4, 6, 9]]

After sorting by end time (events[i][1]):

sorted_events = [[1, 3, 5], [2, 5, 7], [4, 6, 9]]

Sorting takes O(n log n) time, where n is the number of events.

🔑 Step 2: Use Binary Search to Find Compatible Events

For a given event, we need to find the latest event that does not overlap with it.

• Specifically, we want an event whose end time is less than the start time of the current event.
• To do this efficiently, we use binary search on a list of previously processed event end times.

Why Binary Search?

Binary search reduces the complexity of finding compatible events to O(log n) per query.

• Instead of scanning through all prior events to check for compatibility, binary search allows us to directly jump to the relevant candidate.

We use the Python bisect_right function, which returns the position where the current event’s start time would fit in the list of previously processed event end times.

🔑 Step 3: Maintain Rolling Maximum Values

To keep track of the maximum value of previously processed events, we maintain two lists:

1. max_weights: Stores the maximum values observed so far.
   • At each step, this list tells us the best value we can achieve by attending only one event up to the current point.
2. max_weight_ends: Stores the corresponding end times for the values in max_weights.
   • This allows us to check compatibility efficiently using binary search.

Updating max_weights and max_weight_ends:

• After processing each event, we compare its value with the current maximum in max_weights.
• If the new event has a higher value, we append it to max_weights along with its end time to max_weight_ends.

Example:
For events:

events = [[1, 3, 5], [4, 6, 10], [7, 9, 15]]

1. Start with:

   max_weights = [0]  # Rolling maximum values
   max_weight_ends = [-1]  # Corresponding end times
   

2. After processing [1, 3, 5]:

   max_weights = [0, 5]
   max_weight_ends = [-1, 3]
   

3. After processing [4, 6, 10]:

   max_weights = [0, 5, 10]
   max_weight_ends = [-1, 3, 6]
   

4. After processing [7, 9, 15]:

   max_weights = [0, 5, 10, 15]
   max_weight_ends = [-1, 3, 6, 9]
   

🔑 Step 4: Calculate the Maximum Sum

For each event, calculate the potential maximum value by combining:

1. The value of the current event.
2. The maximum value of a compatible previous event (found using binary search).

If the current event’s start time is start and value is value:

• Use bisect_right(max_weight_ends, start - 1) to find the latest compatible event.
• Add its value from max_weights.
• Update the global maximum (max_two) if this combination is better.

🔑 Example Walkthrough

Let’s process a more detailed example:

Input:

events = [[1, 5, 3], [6, 10, 10], [11, 15, 20], [16, 20, 30]]

Step-by-Step Execution:

1. Sort Events by End Time:

   sorted_events = [[1, 5, 3], [6, 10, 10], [11, 15, 20], [16, 20, 30]]
   

2. Initialize Trackers:

   max_weights = [0]  # Initial maximum value
   max_weight_ends = [-1]  # Initial end times
   max_two = 0  # Final result
   

3. Process Events:

   • Event [1, 5, 3]:
     • No compatible previous event (bisect_right([-1], 0) - 1 = -1).
     • max_two = max(0, 3 + 0) = 3.
     • Update trackers:

       max_weights = [0, 3]
       max_weight_ends = [-1, 5]
       

   • Event [6, 10, 10]:
     • Compatible event: [1, 5, 3] (bisect_right([5], 5) - 1 = 0).
     • max_two = max(3, 10 + 3) = 13.
     • Update trackers:

       max_weights = [0, 3, 10]
       max_weight_ends = [-1, 5, 10]
       

   • Event [11, 15, 20]:
     • Compatible event: [6, 10, 10] (bisect_right([10], 10) - 1 = 1).
     • max_two = max(13, 20 + 10) = 30.
     • Update trackers:

       max_weights = [0, 3, 10, 20]
       max_weight_ends = [-1, 5, 10, 15]
       

   • Event [16, 20, 30]:
     • Compatible event: [11, 15, 20] (bisect_right([15], 15) - 1 = 2).
     • max_two = max(30, 30 + 20) = 50.
     • Update trackers:

       max_weights = [0, 3, 10, 20, 30]
       max_weight_ends = [-1, 5, 10, 15, 20]
       

4. Final Result:

   max_two = 50
   

🐍 Python Code

from bisect import bisect_right

class Solution:
    def maxTwoEvents(self, events: List[List[int]]) -> int:
        # Step 1: Pre-sort events by their ending times
        events.sort(key=lambda x: x[1])
        
        # Step 2: Initialize trackers for maximum values and ends
        max_weights = [0]  # Rolling max weights
        max_weight_ends = [-1]  # Rolling end times
        
        max_two = 0  # Final result
        
        # Step 3: Iterate over each event
        for start, end, value in events:
            # Find the max-weight event ending before the current event's start time
            index = bisect_right(max_weight_ends, start - 1) - 1
            
            # Update max_two with the best sum of values
            if value + max_weights[index] > max_two:
                max_two = value + max_weights[index]
            
            # Update rolling max values
            if value > max_weights[-1]:
                max_weights.append(value)
                max_weight_ends.append(end)
        
        return max_two

🔍 Example Walkthrough (With Higher Numbers)

Input:

events = [[1, 5, 10], [6, 10, 15], [11, 15, 20], [16, 20, 25]]

Sorted Events:

[[1, 5, 10], [6, 10, 15], [11, 15, 20], [16, 20, 25]]

Step-by-Step Execution:

1. First Event: [1,5,10]
   • No previous non-overlapping event.
   • max_two = max(0, 10) = 10.
   • Update max_weights = [0, 10], max_weight_ends = [-1, 5].
2. Second Event: [6,10,15]
   • Previous compatible event: [1,5,10].
   • max_two = max(10, 15 + 10) = 25.
   • Update max_weights = [0, 10, 15], max_weight_ends = [-1, 5, 10].
3. Third Event: [11,15,20]
   • Previous compatible event: [6,10,15].
   • max_two = max(25, 20 + 15) = 35.
   • Update max_weights = [0, 10, 15, 20], max_weight_ends = [-1, 5, 10, 15].
4. Fourth Event: [16,20,25]
   • Previous compatible event: [11,15,20].
   • max_two = max(35, 25 + 20) = 45.
   • Update max_weights = [0, 10, 15, 20, 25], max_weight_ends = [-1, 5, 10, 15, 20].

Output:

45

⏱️ Time Complexity

Sorting:

Sorting the events takes O(n log n), where n is the number of events.

Binary Search:

For each event, finding the best compatible event takes O(log n) using binary search.

Overall:

O(n log n), which is efficient for the given constraints.

🔚 Conclusion

This problem elegantly combines sorting, binary search, and dynamic programming for an efficient solution. 🧠 By leveraging a clear structure and precomputed results, we ensure that we can handle large inputs efficiently.

Now, you’re ready to tackle event optimization problems with confidence! 💪 Keep exploring, keep optimizing, and happy coding! 🚀