Question

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

Note:

You may assume the interval’s end point is always bigger than its start point. Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.

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.


Example 2:

Input: [ [1,2], [1,2], [1,2] ]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.


Example 3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

Solution

  • Method 1: Sort + Greedy 
    class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        int len = intervals.length;
        if(len == 0) return 0;
        Arrays.sort(intervals, (a, b)->{
            return a[1] - b[1];
        });
        int end = intervals[0][1];
        int count = 1;
        for(int i = 1; i < len; i++){
            if(intervals[i][0] >= end){
                end = intervals[i][1];
                count++;
            }
        }
        return len - count;
    }
}