Formatted question description: https://leetcode.ca/all/435.html

435. Non-overlapping Intervals (Medium)

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.

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.

 

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.

Related Topics:
Greedy

Similar Questions:

Solution 1. Interval Scheduling Maximization Problem

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

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        int ans = 0, e = INT_MIN;
        for (auto &v : A) {
            if (v[0] >= e) e = v[1];
            else ++ans;
        }
        return ans;
    }
};

Java

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals == null || intervals.length == 0)
            return 0;
        Arrays.sort(intervals, new Comparator<int[]>() {
            public int compare(int[] interval1, int[] interval2) {
                if (interval1[0] != interval2[0])
                    return interval1[0] - interval2[0];
                else
                    return interval1[1] - interval2[1];
            }
        });
        int eraseCount = 0;
        int prev = 0;
        int length = intervals.length;
        for (int i = 1; i < length; i++) {
            if (intervals[prev][1] > intervals[i][0]) {
                if (intervals[prev][1] > intervals[i][1])
                    prev = i;
                eraseCount++;
            } else
                prev = i;
        }
        return eraseCount;
    }
}

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