3. 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.

Note that intervals which only touch at a point are non-overlapping. For example, [1, 2] and [2, 3] are 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

Approach - Greedy (sort start)

• we just maintain the previous end and then check if our new lies in before that meaning we can remove that then we update the previous end
• Time & Space Complexity
  Time complexity: O(nlog⁡n)
  Space complexity: O(1) or O(n) depending on the sorting algorithm.

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a,b) -> a[0] - b[0]);
        int prevEnd = intervals[0][1], res = 0;
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= prevEnd)
                prevEnd = intervals[i][1];
            else {
                res++;
                prevEnd = Math.min(intervals[i][1], prevEnd);
            }
        }
        return res;
    }
}