1288. Remove Covered Intervals

Description

Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list.

The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d.

Return the number of remaining intervals.

 

Example 1:

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2:

Input: intervals = [[1,4],[2,3]]
Output: 1

 

Constraints:

  • 1 <= intervals.length <= 1000
  • intervals[i].length == 2
  • 0 <= li < ri <= 105
  • All the given intervals are unique.

Solutions

Solution 1

Python Code
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class Solution:
    def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: (x[0], -x[1]))
        cnt, pre = 1, intervals[0]
        for e in intervals[1:]:
            if pre[1] < e[1]:
                cnt += 1
                pre = e
        return cnt

Java Code
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class Solution {
    public int removeCoveredIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0] == 0 ? b[1] - a[1] : a[0] - b[0]);
        int[] pre = intervals[0];
        int cnt = 1;
        for (int i = 1; i < intervals.length; ++i) {
            if (pre[1] < intervals[i][1]) {
                ++cnt;
                pre = intervals[i];
            }
        }
        return cnt;
    }
}

C++ Code
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class Solution {
public:
    int removeCoveredIntervals(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b) { return a[0] == b[0] ? b[1] < a[1] : a[0] < b[0]; });
        int cnt = 1;
        vector<int> pre = intervals[0];
        for (int i = 1; i < intervals.size(); ++i) {
            if (pre[1] < intervals[i][1]) {
                ++cnt;
                pre = intervals[i];
            }
        }
        return cnt;
    }
};

Go Code
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func removeCoveredIntervals(intervals [][]int) int {
	sort.Slice(intervals, func(i, j int) bool {
		if intervals[i][0] == intervals[j][0] {
			return intervals[j][1] < intervals[i][1]
		}
		return intervals[i][0] < intervals[j][0]
	})
	cnt := 1
	pre := intervals[0]
	for i := 1; i < len(intervals); i++ {
		if pre[1] < intervals[i][1] {
			cnt++
			pre = intervals[i]
		}
	}
	return cnt
}