LeetCode-in-Java

435. Non-overlapping Intervals

Medium

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

Solution

import java.util.Arrays;

public class Solution {
    /*
     * This is sorting my starting time, the key here is that we'll want to update end time when an
     * erasure is needed: we use the smaller end time instead of the bigger one which is more likely
     * to overlap with others.
     */
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        int erasures = 0;
        int end = intervals[0][1];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < end) {
                erasures++;
                end = Math.min(end, intervals[i][1]);
            } else {
                end = intervals[i][1];
            }
        }
        return erasures;
    }
}

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