# LC435 - Non-overlapping Intervals

## Problem

Given an array of intervals `intervals` where `intervals[i] = [starti, endi]`, return *the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping*.

### Example

**Input:** intervals = `[[1,2],[2,3],[3,4],[1,3]]`

**Output:** 1

**Explanation:** `[1,3]` can be removed and the rest of the intervals are non-overlapping.

## Solution

Time complexity is $$O(n\log n)$$. Space complexity is $$O(1)$$.

```python
def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
	intervals.sort()

	res = 0	
	prev = intervals[0][1]

	for start, end in intervals[1:]:
		# not overlapping
		if start >= prev:
			prev = end
			
		# overlapping
		else: 
			res += 1
			prev = min(end, prev)
	return res
		
```


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