Minimum number of platforms required for a railway station Challenge
Given daily arrival and departure times of every train that reaches a railway station, find the minimum number of platforms required for the railway station so that no train waits.
In other words, find the maximal number of trains simultaneously present in the station.
Input

a pair of lists of times: arrivals and departures; the two lists have same length; arrival i corresponds to the same train as departure i.
alternatively, a list of pairs of times, or any equivalent.
times are numbers between 0, included, and 24, excluded.
there are no dates, only times: input is the daily schedule and repeats every day.
the departure time of a train can be lower than its arrival time; in that case, the train is understood to arrive on a day and depart on the next day; that train will require a platform before midnight and after midnight.
if the arrival time is lower than the departure time, the train is understood to arrive and depart on the same day.
input can be restricted to integers

Output

one integer, the minimum required number of platforms.

Test cases

arrivals   = [10, 13, 16]
departures = [12, 15, 18]
out = 1

arrivals   = [10, 11]
departures = [12, 13]
out = 2

arrivals   = [ 1, 3, 7, 9,10,10,19,23]
departures = [11, 4,11,10,11, 2, 2, 2]
out = 5

arrivals   = [1, 2]
departures = [2, 3]
out = 2

arrivals   = [1, 2]
departures = [3, 2]
out = 2

arrivals   = [2, 22]
departures = [5,  6]
out = 2

Rules

This is code-golf, the shortest code in bytes wins!

Related challenges

Count the timespans
Python, 62 bytes

lambda l:max(sum(a-b^b-h^h-a&lt;1for a,b in l)for h in range(24))

Try it online!
Golfing Surculose Sputum's solution. The new part is a-b^b-h^h-a&lt;1 to check whether the time h lies in the interval from a to b taken cyclically, that is, their sorted ordering is a cyclic permutation of [a,h,b]. To do this, we check whether an odd number of the differences a-b, b-h,h-a are negative. I first did this with multiplication (a-b)^(b-h)^(h-a)&lt;1. But, xor (^) lets us do the same and cuts parens with better precedence.