2064

2064. Minimized Maximum of Products Distributed to Any Store (Medium)

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 be 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

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Solution 1. Binary Search

Intuition:

1. Given k (the maximum number of products given to any store), we can compute the number of stores we are able to assign in O(Q) time where Q is the length of quantities.
2. k is in range [1, MAX(quantities)] and this answer range has perfect monotonicity: There must exist a K that for every k >= K, we can do the assignment with k; for every k < K, we can't do the assignment because the number of stores are not enough.

So, we can use binary search to find this K in O(Qlog(MAX(quantities))) time.

// OJ: https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/
// Author: github.com/lzl124631x
// Time: O(QlogM) where `Q` is the length of `quantities` and `M` is the max element in `quantities`.
// Space: O(1)
class Solution {
public:
    int minimizedMaximum(int n, vector<int>& Q) {
        long long L = 1, R = accumulate(begin(Q), end(Q), 0LL);
        auto valid = [&](int M) {
            int ans = 0;
            for (int n : Q) ans += (n + M - 1) / M; // ceil(n / M)
            return ans <= n;
        };
        while (L <= R) {
            long long M = (L + R) / 2;
            if (valid(M)) R = M - 1;
            else L = M + 1;
        }
        return L;
    }
};

Or use L < R template

// OJ: https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/
// Author: github.com/lzl124631x
// Time: O(QlogM) where `Q` is the length of `quantities` and `M` is the max element in `quantities`.
// Space: O(1
class Solution {
public:
    int minimizedMaximum(int n, vector<int>& Q) {
        long long L = 1, R = *max_element(begin(Q), end(Q));
        auto valid = [&](int M) {
            int ans = 0;
            for (int n : Q) {
                ans += (n + M - 1) / M;
            }
            return ans <= n;
        };
        while (L < R) {
            long long M = (L + R) / 2;
            if (valid(M)) R = M;
            else L = M + 1;
        }
        return L;
    }
};

Discuss

https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1563749/C%2B%2B-Binary-Search