I have tried something similar (but less elegant), however this algorithm does not pass the test with the dataset:

customers {100, 90, 90, 60, 50}
n 2

this algorithm return 210 (100+60+50 = 210 vs 90 + 90 =180)
but true result: 200 (100 + 90 = 190 vs 90 + 60 + 50 = 200)

correct me, if i am wrong. This solution takes n^2 running time due to for each customer, this program is finding the register by checking each register's least. Space-wise, it is O(n) where the n is the number of registers.

There are specifications:

input customers: an array of positive integers representing the queue.

So a (silly and absurd) input of a negative queue will never happen.

Very nice! However, it will crash when (n <= 0).

it is so beautiful and I am superised that my code is similar to this solusion ,I'm just a noob guy here XD

Hey @mildlyeastern,

That's a great question! It's referred to as a ternary. It's a one line conditional statement where the condition is followed the truthy evaluation followed by the else. So like this:

condition == true ? ifConditionWasTrue : elseIfItWasNot

aka

if (condition == true) {
return ifConditionWasTrue;
}else{
return elseIfItWasNot
}

Can someone explain to me what the ? and the : does in the return statement.

For those unfamiliar, the problem is also known as "Bin Packing". I was very suprised but the simplicity of solution, when I first heard about it as well :D

If you have 20 customers and 1 guy - iterations of outer and inner are 20.
if you have 5 customers and 100 guys - iterations of outer and inner are 500.

So simply O(customers.size() * n).

The thing you have to note here is that n is "int", so the solution is more than acceptable.

This is obviously a very elegant solution but wouldn't this be O(n^2)?

4 lines? kill myself

I just want to comment on the fact that I gasped audibly when I understood what you were doing here. Very elegant and simple solution to the problem.