1229. Meeting Scheduler

interval

Given the availability time slots arrays slots1 and slots2 of two people and a meeting duration duration, return the earliest time slot that works for both of them and is of duration duration.

If there is no common time slot that satisfies the requirements, return an empty array.

The format of a time slot is an array of two elements [start, end] representing an inclusive time range from start to end.

It is guaranteed that no two availability slots of the same person intersect with each other. That is, for any two time slots [start1, end1] and [start2, end2] of the same person, either start1 > end2 or start2 > end1.

Example 1:

Input: slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 8
Output: [60,68]

Example 2:

Input: slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 12
Output: []

Constraints:

• 1 <= slots1.length, slots2.length <= 104

• slots1[i].length, slots2[i].length == 2

• slots1[i][0] < slots1[i][1]

• slots2[i][0] < slots2[i][1]

• 0 <= slots1[i][j], slots2[i][j] <= 109

• 1 <= duration <= 106

分析

• 对两个slot数组排好序

• 双指针遍历，找有重叠的部分 [max(s1, s2), min(e1, e2)]

• 如果重叠长度够长（end - start >= duration），返回 [start, start + duration]

• 小的时间段指针往后移

class Solution:
    def minAvailableDuration(self, slots1: List[List[int]], slots2: List[List[int]], duration: int) -> List[int]:
        n,m = len(slots1), len(slots2)
        slots1.sort()
        slots2.sort()
        i = j = 0
        while i < n and j < m:
            s1, e1 = slots1[i]
            s2, e2 = slots2[j]
            
            start = max(s1, s2)
            end = min(e1, e2)
            if end - start  >= duration:
                return [start, start + duration]
            
            if e2 > e1:
                i += 1
            elif e1 > e2:
                j += 1
            else:
                i += 1
                j += 1
        return []