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

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.

I really like this solution.

Shows me that I included one too many steps, utilising min_element AND distance to retrieve the index of the next smallest element.

how to take 16 lines of code and turn it into 5 ;)
very well done, nice to see how smooth it can be.
also best/fastest solution due to use of min/max ^^

beautiful codes!

I think I get it. Thank you

The idea is to "simulate" all the rows with an easy method: just take the numbers of customers and add to the queue with the lowest value. Then you return the highest value queue.

in detail: first we create another vector of values of fixed size (n) and set the values to 0.
then we iterate all the customers values (int i : customers) where i is getting the value of customers[0],[1]...etc; and we add that value to de lowest value queue.

at the end we got a vector of n values (queues) and we just have to return the max value of them.