MEDIUM NC#132 Blind #37 Intervals / Greedy
435. Non-overlapping Intervals
📖 Problem
Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Greedy reasoning pattern
💡 Approach
- → Sort intervals by end time
- → Count non-overlapping intervals greedily
- → If current start < last end, they overlap
- → Time: O(n log n), Space: O(1)
🛠️ Hints & Pitfalls
Hints
- •Sort intervals by end time
- •Count non-overlapping intervals greedily
- •If current start < last end, they overlap
Common Pitfalls
- •Time: O(n log n), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
eraseOverlapIntervals([[1,2],[2,3],[3,4],[1,3]]); Expected:
1 Test Case 2
Not run Input:
eraseOverlapIntervals([[1,2],[1,2],[1,2]]); Expected:
2 Test Case 3
Not run Input:
eraseOverlapIntervals([[1,2],[2,3]]); Expected:
0 📝 Code Editor
📤 Output