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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.
But you need one vector of each size anyway. Here the vectors are moved to the result, in your solution they are copied instead, so you have one extra vector in the end and copying instead of filling with subsequent moving.
This is not best practice. With each level you are filling a new vector. Why not just keep a global sub vector that you increment by one element each iteration?
This is obviously a very elegant solution but wouldn't this be O(n^2)?
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Could you fix the issue above?
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