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435. Non-overlapping Intervals

Description

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

Solutions

Greedy.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• class Solution {
  public:
      int eraseOverlapIntervals(vector<vector<int>>& intervals) {
          sort(intervals.begin(), intervals.end(), [](const auto& a, const auto& b) { return a[1] < b[1]; });
          int ans = 0, t = intervals[0][1];
          for (int i = 1; i < intervals.size(); ++i) {
              if (t <= intervals[i][0])
                  t = intervals[i][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
  
  

• 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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