While this has many up-votes for best practices, it's worth noting that searching the entire list of queues for every single customer results in a worst case time complexity of O(nm), where n is the number of queues and m is the number of customers. You can achieve O(n log m) using Python's heapq module.

Oh my God.

As several people have upvoted this as best practice, it's important to provide the counterpoint.

This solution is O(4n) when it's trivial to write an O(n) solution that's easier to understand.

This code is convoluted for what can be an easy to understand set of if statements in approximately the same number of LOC.

Note that this has time complexity O(N^2) while it's possible to achieve O(N).

This handles punctuation

You're right: most solutions here are O(n^2), but this one is O(n). The code duplication makes it read like a C program, so initially I was surprised.

It doesn't modify the arrays, but creates a new array.

Just a tip: comments should explain what is not obvious from the code, otherwise they end up being a burden.