Time Planner 🎻

Question

Implement a function that given the availability, and , of two people and a meeting duration , returns the earliest time slot that works for both of them and is of duration . If there is no common time slot that satisfies the duration requirement, return an empty array.

Time is given in a Unix format called Epoch, which is a nonnegative integer holding the number of seconds that have elapsed since 00:00:00 UTC, Thursday, 1 January 1970.

Each person’s availability is represented by an array of pairs. Each pair is an epoch array of size two. The first epoch in a pair represents the start time of a slot. The second epoch is the end time of that slot. The input variable is a positive integer that represents the duration of a meeting in seconds. The output is also a pair represented by an epoch array of size two.

In your implementation assume that the time slots in a person’s availability are disjointed, i.e, time slots in a person’s availability don’t overlap. Further, assume that the slots are sorted by slots’ start time.

Implement an efficient solution and analyze its time and space complexities.

Examples:

input:

slotsA = [[10, 50], [60, 120], [140, 210]]

slotsB = [[0, 15], [60, 70]]

dur = 8

output: [60, 68]

input:

slotsA = [[10, 50], [60, 120], [140, 210]]

slotsB = [[0, 15], [60, 70]]

dur = 12

output: [] # since there is no common slot whose duration is 12

Solution

Time complexity O(n²)

Space complexity O(1)

Time complexity O(n+m), where n is the size of the “a” array and m size of the “b” array.

Space complexity O(1)

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