# Time Planner đť

**Question**

Implement a function `meetingPlanner`

that given the availability, `slotsA`

and `slotsB`

, of two people and a meeting duration `dur`

, returns the *earliest* time slot that works for both of them and is of duration `dur`

. 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 personâs 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 `dur`

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 personâs availability are disjointed, i.e, time slots in a personâs availability donât overlap. Further, assume that the slots are sorted by slotsâ 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(nÂ˛)Space complexity

O(1)

Time complexity

O(n+m), where n is the size of the âaâ array and m size of the âbâ array.Space complexity

O(1)