# LeetCode Non-overlapping Intervals

LeetCode 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.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

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.```

```bool comp(const Interval &i1, const Interval &i2){
return i1.start < i2.start || (i1.start == i2.start && i1.end < i2.end);
}

class Solution {
public:
int eraseOverlapIntervals(vector<Interval>& intervals) {
if(intervals.empty()) return 0;
sort(intervals.begin(), intervals.end(), comp);
int ans = 1, n = intervals.size(), last_end = intervals[0].end;
for(int i = 1; i < n; ++i){
if(intervals[i].start >= last_end){
++ans;
last_end = intervals[i].end;
} else {
last_end = min(last_end, intervals[i].end);
}
}
return n - ans;
}
};
```