# 435. Non-overlapping Intervals

## Problem

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

Example 1:

```
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.
```

Example 2:

```
Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
```

Example 3:

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

## Code

```
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
int[] head = intervals[0];
int res = 0;
for(int i = 1; i < intervals.length; i++) {
int[] curr = intervals[i];
if(curr[0] < head[1]) {
res++;
head[1] = Math.min(head[1], curr[1]); // [[1,11],[2,100],[3,12],[11,22]]
} else {
head = curr;
}
}
return res;
}
}
```

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436. Find Right Interval