Pattern visualizer

Non-overlapping Intervals

The insight: to keep the most intervals — and therefore remove the fewest — always keep whichever conflicting interval ends soonest. An interval that ends earlier leaves more room for everything that comes after it, so it's never a worse choice than one that ends later. Sorting by end time and greedily accepting any interval that starts at or after the last kept interval's end applies that rule automatically, so whatever gets skipped along the way is exactly the minimum set that had to be discarded. Sorted order: [[1,2], [2,3], [1,3], [3,4]] with ends 2,3,3,4 (tie at end=3 keeps [2,3] before [1,3]); minimum removals = 1. Animated on: Given an array of intervals, find the minimum number of intervals to remove to make the rest non-overlapping..