[leetcode] 435.Non-overlapping Intervals

Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals
you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are
non-overlapping.

Logic

Sort array's elements by their end and delete sorted elements whose start are less than previous one's end.

Time Complexity

O(n)

Space Complexity

O(n)

Code

public class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals.length == 0)
            return 0;
        PriorityQueue<Interval> pq = new PriorityQueue<Interval>(10, new MyComparator());
        for(Interval it: intervals)
            pq.add(it);
        
        
        int endpoint = 0;
        int count = 0;
        while(pq.size()>1){
            endpoint = pq.poll().end;
            while(pq.size()>0&&pq.peek().start<endpoint){
                pq.poll();
                count++;
            }
        }
        
        return count;
    }
    
    public class MyComparator implements Comparator<Interval>{
        public int compare(Interval a, Interval b){
            
            return a.end - b.end;
        }
        
        
    }
}