### LeetCode Challenge #435. Non-overlapping Intervals

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 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.
```

Constraints:

• `1 <= intervals.length <= 105`
• `intervals[i].length == 2`
• `-5 * 104 <= starti < endi <= 5 * 104`
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
sort(intervals.begin(),intervals.end());
int overlaps=0;
pair<int,int> range;
range.first=intervals[0][0];
range.second=intervals[0][1];

int i=1;

while(i<intervals.size())
{
if(range.second>intervals[i][0])
{
overlaps++;
if(range.second>intervals[i][1])
{
range.first=intervals[i][0];
range.second=intervals[i][1];
}
}
else{
range.first=intervals[i][0];
range.second=intervals[i][1];
}
i++;
}
return overlaps;
}
};
```
```