[LeetCode] 435. 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.
Note:
  1. You may assume the interval's end point is always bigger than its start point.
  2. 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.

Thought process:

The idea is to sort the array based on the intervals start points and use a greedy approach.

After the array is sorted, iterate through the array and use a local variable "end" to keep track of the smallest end point of current overlapping intervals. The reason is that we want to delete all overlapping intervals except the one that has the smallest end point, so that the next interval won't overlap with it.

Once we iterate through a group of overlapping intervals, we update "end" to be the end point of the next interval.

Solution:


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/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals.length < 2) {
            return 0;
        }
        
        Arrays.sort(intervals, new Comparator<Interval>(){
            public int compare(Interval i1, Interval i2) {
                return i1.start - i2.start;
            }
        });
        
        int end = intervals[0].end;
        int min = 0;
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start < end) {
                end = Math.min(end, intervals[i].end);
                min++;
            } else {
                end = intervals[i].end;
            }
        }
        
        return min;
    }
}

Time complexity:

Sorting takes O(nlogn). For loop takes O(n). The overall time complexity is O(nlogn).
Location: Philadelphia, PA, USA

Comments

  1. dhanviJune 17, 2020 at 3:38 AM

    we need to sort according to end time and not start time to allow maximum non-overlapping intervals to be taken

    ReplyDelete
  2. Eman QurashiDecember 24, 2021 at 11:23 PM

    I found that solution is very popular and helpful: https://youtu.be/2NN6N-tNyag

    ReplyDelete

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