# 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.

Example 1:

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.

Example 2:

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.

Example 3:

Input: [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

# Solution

Approach 1: sort by the start of each interval, and greedily remove the longer interval when there's an overlap

# Code (Python)

Approach 1:

    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[0] >= latest:
                latest = interval[1]
                num_removed += 1
                latest = min(latest, interval[1])
        return num_removed

# Code (C++)

Approach 1:

Approach 2: