2188. Minimized Maximum Of Products Distributed To Any Store¶

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2188. 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 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

Solution¶

class Solution {
    public int minimizedMaximum(int n, int[] quantities) {
        int m = quantities.length;
        int maxi = 0;
        for (int ele : quantities) maxi = Math.max(maxi, ele);
        int low = 1;
        int high = maxi;
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (ok(mid, quantities, n)) {
                ans = mid;
                low = mid + 1;
            }
            else high = mid - 1;
        }
        return low;
    }

    private boolean ok(int mid, int arr[], int n) {
        int sum = 0;
        for (int ele : arr) sum += (ele + mid - 1) / mid;
        return sum > n;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here