Non-overlapping Intervals – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Non-overlapping Intervals LeetCode Solution

bool comp(vector<int> &a,vector<int> &b) {
	return a[1]<b[1];
}
class Solution {
public:
	int eraseOverlapIntervals(vector<vector<int>>& intervals) {
		int ans=-1;      
		if(intervals.size()==0) return 0;       
		sort(intervals.begin(),intervals.end(),comp);      //custom comperator is used.
		vector<int> prev= intervals[0];

		for(vector<int> i: intervals) {
			if(prev[1]>i[0]) {
				ans++;                //we dont update previous, because i[1] will be grater then prev[1]
			}else prev=i;           // we want the end point to be minimum
		}
		return ans;                 //ans was initially made -1 because our prev and intervals[0] will always match
	}
};

Java Non-overlapping Intervals LeetCode Solution

 public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals.length == 0)  return 0;

        Arrays.sort(intervals, new myComparator());
        int end = intervals[0].end;
        int count = 1;        

        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start >= end) {
                end = intervals[i].end;
                count++;
            }
        }
        return intervals.length - count;
    }
    
    class myComparator implements Comparator<Interval> {
        public int compare(Interval a, Interval b) {
            return a.end - b.end;
        }
    }
 

Python 3 Non-overlapping Intervals LeetCode Solution

def eraseOverlapIntervals(intervals):
	end, cnt = float('-inf'), 0
	for s, e in sorted(intervals, key=lambda x: x[1]):
		if s >= end: 
			end = e
		else: 
			cnt += 1
	return cnt

Array-1180
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