0435. Non-overlapping Intervals

https://leetcode.com/problems/non-overlapping-intervals

Description

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

ac: Greedy

class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        // edge cases
        if (intervals.length <= 1) return 0;
        
        Arrays.sort(intervals, (a, b) -> {return a.start != b.start ? a.start - b.start : a.end - b.end;});
        
        Interval prev = intervals[0];
        int count = 0;
        for (int i = 1; i < intervals.length; i++) {
            Interval curr = intervals[i];
            if (curr.start < prev.end) { // overlap
                count++; // have to remove one
                prev = curr.end > prev.end ? prev : curr;
            } else {
                prev = curr;
            }
        }
        
        return count;
    }
}
// O(NlogN) time.
// Iterate, if there are 2 intervals overlap, we have to remove one. Which one? The one with larger endpoints, because it has more possibility to overlap with others.