Minimum Removals for No Overlap
Greedy keep-by-earliest-end scheduling.
Minimum Removals for No Overlap is a free Coding Interview Prep lesson on CoddyKit — lesson 4 of 4. You can read the complete lesson below for free — then practise it hands-on in the browser with a built-in code editor and a 24/7 AI tutor. It is part of the Coding Interview Prep learning path, one of 4 lessons in the course, and your progress syncs across the web and the CoddyKit app.
The Removal Goal
You have overlapping intervals and want the fewest removals so none overlap anymore. Keep as many as you can. ✂️
Flip the Problem
Removing the fewest is the same as keeping the most non-overlapping intervals. Solve the keep version, then the removals are n minus kept.
This Is Activity Selection
Keeping the most non-overlapping intervals is the classic activity selection problem in disguise. The same greedy idea solves both.
Sort by End
Here the winning order is by end time, not start. Finishing early frees the timeline soonest for the next interval you might keep.
intervals.sort(key=lambda x: x[1])The Greedy Choice
Always keep the interval that ends earliest among those still compatible. It leaves the maximum room for the rest.
Track the Last Kept End
Hold the end of the last interval you kept. The next interval is compatible only if its start is at or after that boundary.
if start >= last_end:
last_end = endCount the Removals
When an interval starts before last_end, it conflicts, so you drop it and add one to your removal count. Otherwise you keep it.
else:
removed += 1Why Earliest End Wins
An exchange argument proves it: swapping any kept interval for the earliest-ending compatible one never reduces how many you can keep.
Handle the Touching Edge
Decide whether [1, 2] and [2, 3] count as overlapping. If sharing only an endpoint is allowed, use start >= last_end as your test.
The Full Greedy
Sort by end, sweep once, and count conflicts. The total cost is O(n log n) from the sort plus a single linear pass.
removed = 0; last_end = float('-inf')
for s, e in intervals:
if s >= last_end: last_end = e
else: removed += 1A Familiar Shape
This pattern schedules the most meetings in one room or packs the most jobs on one machine. Spot it whenever conflicts must be minimized.
Quick Check
You greedily keep non-overlapping intervals.
Recap
Minimum removals equals n minus the most you can keep. Sort by end, greedily keep earliest-ending compatible intervals, and count the rest. 🚀
Frequently asked questions
Is the “Minimum Removals for No Overlap” lesson free?
Yes — the full text of “Minimum Removals for No Overlap” is free to read here on the web, and the Coding Interview Prep course includes 4 lessons in total. To practise it interactively (a built-in code editor and a 24/7 AI tutor) and unlock the rest of the Coding Interview Prep course, upgrade to CoddyKit PRO.
What will I learn in “Minimum Removals for No Overlap”?
Greedy keep-by-earliest-end scheduling. You practise Coding Interview Prep with hands-on code you run directly in the browser, and a 24/7 AI tutor answers your questions as you work through the lesson.
Do I need any experience to start Coding Interview Prep?
No prior experience is required. Coding Interview Prep on CoddyKit is structured for beginners through advanced learners; this is — lesson 4 of 4, so you can start here or from the beginning and move at your own pace.
How long does the “Minimum Removals for No Overlap” lesson take?
Most CoddyKit lessons take about 5–10 minutes. Each one is bite-sized and interactive, so you make steady progress and pick up exactly where you left off across the web and the app.
Can I write and run code in this Coding Interview Prep lesson?
Yes. Every Coding Interview Prep lesson includes a built-in code editor, so you write and run real code right in your browser and get instant AI feedback — no local setup required.
All lessons in this course
- Sort Intervals by Start
- Merge Overlapping Intervals
- Line Sweep for Max Overlap
- Minimum Removals for No Overlap