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;
        }
    }
    
  • // 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;
        }
    };
    
  • # Definition for an interval.
    # class Interval(object):
    #     def __init__(self, s=0, e=0):
    #         self.start = s
    #         self.end = e
    
    class Solution(object):
      def eraseOverlapIntervals(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        intervals.sort(key=lambda i: i.end)
        ans = 0
        end = float("-inf")
        for interval in intervals:
          # print interval.start, interval.end
          if interval.start >= end:
            ans += 1
            end = interval.end
        return len(intervals) - ans
    
    

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