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2064. Minimized Maximum of Products Distributed to Any Store

Description

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 possible x.

 

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

Binary search.

  • class Solution {
    public int minimizedMaximum(int n, int[] quantities) {
        int left = 1, right = (int) 1e5;
        while (left < right) {
            int mid = (left + right) >> 1;
            int cnt = 0;
            for (int v : quantities) {
                cnt += (v + mid - 1) / mid;
            }
            if (cnt <= n) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

  • class Solution {
public:
    int minimizedMaximum(int n, vector<int>& quantities) {
        int left = 1, right = 1e5;
        while (left < right) {
            int mid = (left + right) >> 1;
            int cnt = 0;
            for (int& v : quantities) {
                cnt += (v + mid - 1) / mid;
            }
            if (cnt <= n) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

  • class Solution:
    def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
        def check(x):
            return sum((v + x - 1) // x for v in quantities) <= n

        return 1 + bisect_left(range(1, 10**6), True, key=check)


  • func minimizedMaximum(n int, quantities []int) int {
	return 1 + sort.Search(1e5, func(x int) bool {
		x++
		cnt := 0
		for _, v := range quantities {
			cnt += (v + x - 1) / x
		}
		return cnt <= n
	})
}

  • function minimizedMaximum(n: number, quantities: number[]): number {
    let left = 1;
    let right = 1e5;
    while (left < right) {
        const mid = (left + right) >> 1;
        let cnt = 0;
        for (const v of quantities) {
            cnt += Math.ceil(v / mid);
        }
        if (cnt <= n) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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