**Problem Statement**

A university has exactly one turnstile. It can be used either as an exit or an entrance. Unfortunately, sometimes many people want to pass through the turnstile and their directions can be different. The ith person comes to the turnstile at time[i] and wants to either exit the university if direction[i] = 1 or enter the university if direciton[i] = 0. People form 2 queues, one to exit and one to enter. They are ordered by the time when they came to the turnstile and, if the times are equal, by their indices.

If some person wants to enter the university and another person wants to leave the university at the same moment, there are three cases:

• If in the previous second the turnstile was not used (maybe it was used before, but not at the previous second), then the person who wants to leave goes first.

• If in the previous second the turnstile was used as an exit, then the person who

wants to leave goes first

• If in the previous second the turnstile was used as an entrance, then the person

who wants to enter goes first

Passing through the turnstile takes 1 second

For each person, find the time when they will pass through the turnstile

The function must return an array of n integers where the value at index[i] is the same when the ith person will pass the turnstile

The function has the following params:

• time: an array of n integers where the value at index i is the time when the

ith person will came to the turnstile

• direction: an array of n integers where the value at index i is the direction

of the ith person

Constraints

• 1 <= n <= 105

• 0 <= time[i] <= 109 for 0 <= i <= n – 1

• time[i] <= time[i+1] for 0 <= i <= n - 2

• 0 <= direction [i] <= 1 for 0 <= i <= n – 1

Example:

**n = 4**

**time = [0,0,1,5]**

**direction = [0,1,1,0]**

**Output = [2,0,1,5]**

Example 2

**n = 5**

time = [0,1,1,3,3]

time = [0,1,1,3,3]

direction = [0,1,0,0,1]

direction = [0,1,0,0,1]

**Output = [0,2,1,4,3]**

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**This is my actual trying**

def getTimes(time, direction) persons = time.size - 1 exits = [] (0..persons).each do |person| if time[person] == time[person + 1] exits[person + 1] = time[person] if direction[person + 1] == 1 and direction[person] == 0 else exits[person] = time[person] if direction[person] != direction[person + 1] end end p exits end

And my response

**[nil, 0, 1, 5]**