How to allocate x copy of n objects among k persons - algorithm

Question

I have following use case;
I have N distinct items , each can have x number of copies. Now I need to distribute these items among k persons where each person's capacity is varying and can be <=N.
Following conditions must be met;
Each person should get one and only one copy of an Item
Example:
Items = apple , banana , orange
copies = 3 ( It means we have 3 apples , 3 bananas and 3 oranges )
So I have a array;
{1,2,3,4,5,6,7,8,9} // 1,2,3 = 3 apples ; 4,5,6 = 3 banana ; 7,8,9 = 3 oranges
Total Person = 5
Person Capacity
P1 3
P2 2
P3 1
P4 1
P5 2
How can I solve such problem ? The problem I am facing is that when I allocate it for an arbitrary numbers for N , x , k , I sometimes end up in a case where I am left with some items to allocate because I can't ensure the condition that "Each person should get one and only one copy of an Item"

Answer 1

Since each item has the same "weight", you can actually solve this problem greedily. To ensure that each person receives distinct items, we create the sequence containing all the xN items by repeating the sequence of N distinct items x times. Then, we go through each of the persons and simply remove and assign them the first c items of this sequence, where c is that person's carrying capacity.
This works because of the way we have laid out the items and because c <= N. In our "mega" sequence, duplicates are always N indices away from each other, so c consecutive elements will never contain two duplicates. Duplicates will only appear in contiguous subsequences containing more than N items.
Note that in the implementation of this algorithm, you don't actually have to create the mega sequence; you can simply repeatedly iterate through the distinct item sequence by using modular arithmetic. To keep the explanation simple, I will be forming the "mega" items sequence in the examples, but you don't have to do this in the implementation.
Taking the example in your question, let the 3 distinct items be A, B, C with 3 copies each. The "mega" sequence is formed by repeating the distinct items sequence 3 times: A, B, C, A, B, C, A, B, C. Now we go through each person and simply assign them the number of items they can carry. To illustrate this, consider the following cases for the capacity array (taken from your question and comments below):
[3, 2, 1, 1, 2]: P1 gets A, B, C, P2 gets A, B, P3 gets C, P4 gets A, and P5 gets B, C.
[2, 2, 2, 2, 1]: P1 gets A, B, P2 gets C, A, P3 gets B, C, P4 gets A, B, and P5 gets C.
[3, 1, 1, 1, 1, 2]: P1 gets A, B, C, P2 gets A, P3 gets B, P4 gets C, P5 gets A, P6 gets B, C.