Minimized Maximum of Products Distributed to

lcpy gen -n 2064   # by problem number
lcpy gen -s minimized_maximum_of_products_distributed_to_any_store   # by problem name

Input: n = 6, quantities = [11,6]
Output: 3
Explanation: One optimal way is:
- The 11 products of type 0 are distributed to the first four stores in these amounts: 2, 3, 3, 3
- The 6 products of type 1 are distributed to the other two stores in these amounts: 3, 3
The maximum number of products given to any store is max(2, 3, 3, 3, 3, 3) = 3.

Input: n = 7, quantities = [15,10,10]
Output: 5
Explanation: One optimal way is:
- The 15 products of type 0 are distributed to the first three stores in these amounts: 5, 5, 5
- The 10 products of type 1 are distributed to the next two stores in these amounts: 5, 5
- The 10 products of type 2 are distributed to the last two stores in these amounts: 5, 5
The maximum number of products given to any store is max(5, 5, 5, 5, 5, 5, 5) = 5.

Input: n = 1, quantities = [100000]
Output: 100000
Explanation: The only optimal way is:
- The 100000 products of type 0 are distributed to the only store.
The maximum number of products given to any store is max(100000) = 100000.

class Solution:
    # Time: O(m * log(max(quantities)))
    # Space: O(1)
    def minimized_maximum(self, n: int, quantities: list[int]) -> int:
        lo, hi = 1, max(quantities)
        while lo < hi:
            mid = (lo + hi) // 2
            if sum(-(-q // mid) for q in quantities) <= n:
                hi = mid
            else:
                lo = mid + 1
        return lo