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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:

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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
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  
  ############
  
  class Solution {
      public int eraseOverlapIntervals(int[][] intervals) {
          Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
          int t = intervals[0][1], ans = 0;
          for (int i = 1; i < intervals.length; ++i) {
              if (intervals[i][0] >= t) {
                  t = intervals[i][1];
              } else {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• // 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;
      }
  };
  

• class Solution:
      def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
          intervals.sort(key=lambda x: x[1])
          ans, t = 0, intervals[0][1]
          for s, e in intervals[1:]:
              if s >= t:
                  t = e
              else:
                  ans += 1
          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
  
  

• func eraseOverlapIntervals(intervals [][]int) int {
  	sort.Slice(intervals, func(i, j int) bool {
  		return intervals[i][1] < intervals[j][1]
  	})
  	t, ans := intervals[0][1], 0
  	for i := 1; i < len(intervals); i++ {
  		if intervals[i][0] >= t {
  			t = intervals[i][1]
  		} else {
  			ans++
  		}
  	}
  	return ans
  }
  

• function eraseOverlapIntervals(intervals: number[][]): number {
      intervals.sort((a, b) => a[1] - b[1]);
      let end = intervals[0][1],
          ans = 0;
      for (let i = 1; i < intervals.length; ++i) {
          let cur = intervals[i];
          if (end > cur[0]) {
              ans++;
          } else {
              end = cur[1];
          }
      }
      return ans;
  }
  
  

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