Greedy earliest finish
Drop the Fewest Reservations
01. The question
Reservations use half-open [start,end) times with start < end. Return the minimum number to discard so one resource can accept all remaining reservations. Ending exactly when another starts is compatible.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input[[1,4],[2,3],[3,5],[5,7]]
Output1
- Keep [2,3] because it finishes first.
- Discard [1,4], which conflicts with that choice.
- Keep [3,5] and [5,7]; touching times require no extra removal.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: [[1,3],[2,4],[3,5]]. Parameter: "". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.
Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Sort by end, then start for deterministic ties.
- Keep each span whose start is at least the last kept end.
- Count each conflicting span as removed.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
def solve(data, parameter=''):
end = None
removed = 0
for start, finish in sorted(data, key=lambda interval: (interval[1], interval[0])):
if end is None or start >= end:
end = finish
else:
removed += 1
return removedHow to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
def solve(data, parameter=''):
end = None
removed = 0
for start, finish in sorted(data, key=lambda interval: (interval[1], interval[0])):
if end is None or start >= end:
end = finish
else:
removed += 1
return removed06. Cost & edge cases
- Time
- O(n log n)
- Space
- O(n) sorting workspace
- Empty input needs zero removals.
- Repeated reservations conflict unless discarded.
- Nested spans favor the earliest end, not the earliest start.