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2064. Minimized Maximum of Products Distributed to Any Store

You are given an integer n indicating there are n specialty retail stores. There are m product types of varying amounts, which are given as a 0-indexed integer array quantities, where quantities[i] represents the number of products of the ith product type.

You need to distribute all products to the retail stores following these rules:

  • A store can only be given at most one product type but can be given any amount of it.
  • After distribution, each store will have been given some number of products (possibly 0). Let x represent the maximum number of products given to any store. You want x to be as small as possible, i.e., you want to minimize the maximum number of products that are given to any store.

Return the minimum possiblex.

Example 1:

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.

Example 2:

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.

Example 3:

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.

Constraints:

  • m == quantities.length
  • 1 <= m <= n <= 105
  • 1 <= quantities[i] <= 105

Solutions (Rust)

1. Solution

implSolution{pubfnminimized_maximum(n:i32,quantities:Vec<i32>) -> i32{letmut low = 1;letmut high = *quantities.iter().max().unwrap();while low < high {let x = (low + high) / 2;letmut y = 0;for q in&quantities { y += q / x;if q % x != 0{ y += 1;}}if y > n { low = x + 1;}else{ high = x;}} high }}
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