1229. Meeting Scheduler π
Description
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
Solutions
Solution 1: Sorting + Two Pointers
We can sort the free time intervals of both people, then use two pointers to traverse the two arrays and find the intersection of the free time intervals of both people. If the length of the intersection is greater than or equal to duration
, return the start time of the intersection and the start time plus duration
. Otherwise, if the end time of the first person's free time interval is less than the end time of the second person's free time interval, move the first person's pointer; otherwise, move the second person's pointer. Continue traversing until a suitable time interval is found or the traversal ends.
The time complexity is $O(m \times \log m + n \times \log n)$, and the space complexity is $O(\log m + \log n)$. Here, $m$ and $n$ are the lengths of the two arrays, respectively.
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