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Minimize Maximum of Products Distributed to Any Store

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You are given an integer n indicating there are n specialty retail stores. There are m product types 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 product to the retail stores following these rules:

  • A store can only be given at most one product type but can 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 of products that are given to any store.

Return the minimum possible x.

Example:

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Input: n = 6, quantities = [11, 6];
Output = 3;
The 11 products of type 0 are distributed to the first four stores in the amount of [2, 3, 3, 3]. The 6 products of type 1 are distributed to the other two stores in the amount of [3, 3].
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Input: n = 7, quantities = [15, 10, 10];
Output: 5;
[5, 5, 5, 5, 5, 5, 5];
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Input: n = 1, quantities = [10000];
Output = 10000;

Constraints:

  • 1 ≤ m ≤ n ≤ 105
  • 1 ≤ quantitiesi ≤ 105

Solution

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class Solution {
public:
    int minimizedMaximum(int n, vector<int>& quantities) {
        int low{1}; // same as 'int low = 1;', but in a more safe way.
        int high = *max_element(quantities.begin(), quantities.end());
        while (low < high) {
            int mid = (low + high) / 2;
            int needed{0};
            for (int q : quantities) {
                needed += (q + mid - 1) / mid;
            }
            if (needed <= n) high = mid;
            else low = mid + 1;
        }
        return low
    }
}

Explanation: The key idea of this solution is binary search, the variable mid here is used to indicate the possible maximum quantity that can be distributed to a single store. Variables low and high hold the lower bound and the upper bound of the quantity for distribution. Then use binary search to update the variable mid to find the possible minimized maximum quantity. The way to update mid is to check whether this value can be distributed within n stores. Here using variable needed to hold the number of store that needs for distributing under mid. So traverse the quantities to compute the needed number of stores to distribute q products under mid is the upper bound. For example, if q=7 then mid=3, since 7 products need 3 stores. If needed is less than n, which means the value of mid is greater, and there might a smaller solution, thus update high with mid. Otherwise, mid is to small, thus update low with mid + 1. When low and high converge, which means the minimized value mid is found, then return low.