Solution Approach

The solution implements the sorting and binary search strategy with precomputation:

Step 1: Sort the events

events.sort()

Events are sorted by start time (first element of each sublist by default). This allows us to use binary search later.

Step 2: Precompute maximum values from each position

n = len(events)
f = [events[-1][2]] * n
for i in range(n - 2, -1, -1):
    f[i] = max(f[i + 1], events[i][2])

• Initialize array f where f[i] will store the maximum value of any single event from index i to the end
• Start with the last event's value for all positions
• Iterate backwards from the second-to-last event, updating f[i] = max(f[i+1], events[i][2])
• This ensures f[i] contains the maximum value achievable from position i onwards

Step 3: Process each event and find the best combination

ans = 0
for _, e, v in events:
    idx = bisect_right(events, e, key=lambda x: x[0])
    if idx < n:
        v += f[idx]
    ans = max(ans, v)

• For each event with end time e and value v:
  • Use bisect_right to find the first event whose start time is greater than e
  • The key=lambda x: x[0] tells bisect to compare only start times
  • bisect_right(events, e, key=lambda x: x[0]) finds the leftmost position where startTime > e
  • If such an event exists (idx < n), add the best value from that position (f[idx]) to the current event's value
  • Update the answer with the maximum value found

Time Complexity: O(n log n) where n is the number of events

• Sorting: O(n log n)
• Precomputation: O(n)
• For each event, binary search: O(log n), total O(n log n)

Space Complexity: O(n) for the precomputed array f

Example Walkthrough

Let's walk through the solution with a small example: events = [[1,3,4], [3,4,3], [5,6,5]]

Step 1: Sort the events by start time After sorting: [[1,3,4], [3,4,3], [5,6,5]] (already sorted in this case)

• Index 0: [1,3,4] - starts at 1, ends at 3, value = 4
• Index 1: [3,4,3] - starts at 3, ends at 4, value = 3
• Index 2: [5,6,5] - starts at 5, ends at 6, value = 5

Step 2: Build the precomputed array f We build f from right to left, where f[i] = maximum value from index i to the end:

• f[2] = 5 (only event at index 2, value = 5)
• f[1] = max(f[2], events[1][2]) = max(5, 3) = 5 (best from index 1 onwards is 5)
• f[0] = max(f[1], events[0][2]) = max(5, 4) = 5 (best from index 0 onwards is 5)

Result: f = [5, 5, 5]

Step 3: Process each event and find best combinations

Processing event [1,3,4] (index 0):

• End time = 3, value = 4
• Binary search: Find first event starting after time 3
  • bisect_right with key on start times finds index 2 (event [5,6,5] starts at 5 > 3)
• Total value = 4 + f[2] = 4 + 5 = 9
• Update answer: ans = max(0, 9) = 9

Processing event [3,4,3] (index 1):

• End time = 4, value = 3
• Binary search: Find first event starting after time 4
  • bisect_right finds index 2 (event [5,6,5] starts at 5 > 4)
• Total value = 3 + f[2] = 3 + 5 = 8
• Update answer: ans = max(9, 8) = 9

Processing event [5,6,5] (index 2):

• End time = 6, value = 5
• Binary search: Find first event starting after time 6
  • bisect_right finds index 3 (no events after, idx = 3 = n)
• Since idx >= n, no valid second event exists
• Total value = 5 (just this event)
• Update answer: ans = max(9, 5) = 9

Final Answer: 9

The maximum value is achieved by attending events [1,3,4] and [5,6,5], which don't overlap and give a total value of 4 + 5 = 9.

Time and Space Complexity

Time Complexity: O(n × log n)

The time complexity is dominated by:

1. Sorting the events array: O(n × log n)

2. Building the suffix maximum array f: O(n) - single pass through the array

3. For each event in the main loop (n iterations):

   • bisect_right operation with custom key: O(log n) binary search
   • Other operations: O(1)

   Total for this loop: O(n × log n)

Overall time complexity: O(n × log n) + O(n) + O(n × log n) = O(n × log n)

Space Complexity: O(n)

The space complexity consists of:

1. The suffix maximum array f: O(n) - stores n elements
2. The sorting operation (Python's Timsort): O(n) in worst case
3. Other variables (ans, idx, loop variables): O(1)

Overall space complexity: O(n)

Where n is the number of events in the input array.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A common pitfall is using a different binary search variant that doesn't match the canonical template. The correct template uses while left <= right with first_true_index tracking.

Incorrect Implementation:

# WRONG: Using while left < right with right = mid
left, right = 0, n
while left < right:
    mid = (left + right) // 2
    if events[mid][0] > end:
        right = mid
    else:
        left = mid + 1
return left

Correct Implementation:

# CORRECT: Using template with first_true_index
left, right = 0, n - 1
first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if events[mid][0] > end:  # feasible condition
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
# Use first_true_index (-1 if not found)

2. Incorrect Overlap Definition

A critical pitfall is misunderstanding when two events overlap. Many developers incorrectly assume that if one event ends at time t and another starts at time t, they don't overlap. However, according to the problem, these events DO overlap - you need the next event to start at t + 1 or later.

Incorrect Implementation:

# WRONG: This would find events where start_time >= end_time
if events[mid][0] >= end:  # Wrong comparison - should be strict >

Correct Implementation:

# CORRECT: Find events where start_time > end_time
if events[mid][0] > end:  # Strict inequality

2. Not Considering Single Event as Answer

Another pitfall is forgetting that attending just one high-value event might be better than attending two lower-value events. The precomputed array handles this, but if you try to optimize and only look for pairs, you'll miss cases where a single event has the maximum value.

Example: Events: [[1,2,10], [4,5,1], [6,7,1]]

• Best two events: value = 10 + 1 = 11
• But if we had [[1,2,20], [4,5,1], [6,7,1]], the single event with value 20 is better than any pair

Solution: The code correctly handles this by initializing current_sum = value (considering the single event) before potentially adding a second event.

3. Off-by-One Error in Precomputation

When building the max_value_from array, starting from the wrong index or using wrong boundaries can cause incorrect results.

Incorrect:

# WRONG: Missing the last event's value
max_value_from = [0] * n
for i in range(n - 2, -1, -1):
    max_value_from[i] = max(max_value_from[i + 1], events[i][2])

Correct:

# CORRECT: Initialize last element first
max_value_from = [0] * n
max_value_from[-1] = events[-1][2]  # Don't forget this!
for i in range(n - 2, -1, -1):
    max_value_from[i] = max(max_value_from[i + 1], events[i][2])

4. Using bisect_left Instead of Custom Binary Search

Using bisect_left or bisect_right without proper understanding can lead to wrong results. The problem requires finding the first event where start_time > end_time, which needs careful implementation.

Why the manual binary search is clearer:

• It explicitly shows we're looking for events[mid][0] > end
• The condition is unambiguous
• Using bisect_right(events, end, key=lambda x: x[0]) would find where to insert end to keep the list sorted by start times, which gives us the first event with start_time > end

5. Not Handling Edge Cases

Forgetting to check if a valid second event exists before accessing the precomputed array.

Incorrect:

# WRONG: May cause index out of bounds
current_sum = value + max_value_from[left]

Correct:

# CORRECT: Check bounds first
current_sum = value
if left < n:
    current_sum += max_value_from[left]