LeetCode 435. Non-overlapping Intervals

Question

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

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

Solution

// sort by start, greedy, Time Complexity O(nlog(n))
public int eraseOverlapIntervals(int[][] intervals) {
    if (intervals == null || intervals.length == 0) {
        return 0;
    }
    Arrays.sort(intervals, new Comparator<int[]>() {
        @Override
        public int compare(int[] o1, int[] o2) {
            return o1[0] - o2[0];
        }
    });
    int numRemove = 0;
    int prevEnd = intervals[0][1];
    for (int i = 1; i < intervals.length; i++) {
        int[] curr = intervals[i];
        if (curr[0] >= prevEnd) { // no overlap
            prevEnd = curr[1];
        } else if (prevEnd < curr[1]) { // intersection overlap
            // remove current interval
            // do not change prevEnd
            numRemove++;
        } else { // included overlap
            // remove previous interval
            numRemove++;
            prevEnd = curr[1];
        }
    }
    return numRemove;
}