Array     Dynamic Programming     Greedy     Medium     Sorting    

Problem Statement:

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

Solution:

bool mycomp(vector<int>&a, vector<int>&b)
{
    return a[1]<b[1];
}
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) 
    {
        if (intervals.size()==1) return 0;
        sort(intervals.begin(), intervals.end(), mycomp);
        int n=intervals.size(), curr=intervals[0][1], res=0;
        for (int i=1; i<n; i++)
        {
            if (intervals[i][0]>=curr) curr = intervals[i][1];
            else res ++;
        }
        return res;
    }
};

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals = sorted(intervals, key=lambda k: k[1])
        n, res, curr = len(intervals), 0, intervals[0][1]
        for i in range(1,n):
            if intervals[i][0]>=curr:
                curr = intervals[i][1]
			else:
			    res += 1
        return res