# 435. Non-overlapping Intervals
Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Input: [[1,2],[2,3],[3,4],[1,3]] Output: 1 Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Input: [[1,2],[1,2],[1,2]] Output: 2 Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Input: [[1,2],[2,3]] Output: 0 Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Approach 1: sort by the start of each interval, and greedily remove the longer interval when there's an overlap
# Code (Python)
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: # idea: greedy -- always remove the longer interval when there's an overlap num_removed = 0 latest = -float('inf') for interval in sorted(intervals): if interval >= latest: latest = interval else: num_removed += 1 latest = min(latest, interval) return num_removed
# Code (C++)