動機

就是merge interval，但要記得紀錄merge幾個interval

Problem

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: 2Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2:

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

Example 3:

Input: intervals = [[0,10],[5,12]]Output: 2

Example 4:

Input: intervals = [[3,10],[4,10],[5,11]]Output: 2

Example 5:

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

 

Constraints:

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

Sol

class Solution:
    def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: (x[0],-x[1]))
        stk = [[intervals[0],1]]
        for c,d in intervals[1:]:
            a,b = stk[-1][0]
            if a <= d <= b:
                continue
            elif a <= c <= b:
                stk[-1][0] = [min(a,c),max(b,d)]
                stk[-1][1] += 1
            else:
                stk.append([[c,d],1])
            #print(stk)
        return sum([x[1] for x in stk])