# 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]
else:
num_removed += 1
latest = min(latest, interval[1])
return num_removed
# Code (C++)
Approach 1:
Approach 2: