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Non-Overlapping Intervals LC #435 Medium Intervals · Greedy
Problem

Given an array of intervals, return the minimum number of intervals to remove so the remaining intervals are non-overlapping.

Example
Input: [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: Remove [1,3] — it overlaps [1,2] and [2,3]. Keep [1,2],[2,3],[3,4].
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Approach
Greedy — Sort by End Time
Keep interval with earliest end (leaves room for more). If curr.start < prevEnd → remove current.
Intervals (Sorted by End)
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Removals 0
Variables
curr
—
prevEnd
—
Overlap?
—
Action
—
Step Logic
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Algorithm
1
Sort by end time
2
If curr.start ≥ prevEnd → keep, update prevEnd
3
Else → remove current (count++), prevEnd unchanged
Time
O(n log n)
Space
O(1)
Why Sort by End

Keeping the interval with the earliest end leaves maximum room for future intervals. When two overlap, we remove the one that ends later (current, since sorted by end).