Formatted question description: https://leetcode.ca/all/435.html

# 435. Non-overlapping Intervals (Medium)

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

**Example 2:**

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

**Example 3:**

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

**Note:**

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

**Related Topics**:

Greedy

**Similar Questions**:

## Solution 1. Interval Scheduling Maximization Problem

```
// OJ: https://leetcode.com/problems/non-overlapping-intervals/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
int ans = 0, e = INT_MIN;
for (auto &v : A) {
if (v[0] >= e) e = v[1];
else ++ans;
}
return ans;
}
};
```

Java

```
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if (intervals == null || intervals.length == 0)
return 0;
Arrays.sort(intervals, new Comparator<int[]>() {
public int compare(int[] interval1, int[] interval2) {
if (interval1[0] != interval2[0])
return interval1[0] - interval2[0];
else
return interval1[1] - interval2[1];
}
});
int eraseCount = 0;
int prev = 0;
int length = intervals.length;
for (int i = 1; i < length; i++) {
if (intervals[prev][1] > intervals[i][0]) {
if (intervals[prev][1] > intervals[i][1])
prev = i;
eraseCount++;
} else
prev = i;
}
return eraseCount;
}
}
```