LeetCode 2064: Minimized Maximum of Products Distributed to Any Store Solution

Master LeetCode problem 2064 (Minimized Maximum of Products Distributed to Any Store), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2064. Minimized Maximum of Products Distributed to Any Store

Medium

Array Binary Search Greedy

Problem Explanation

Explanation:

To minimize the maximum number of products given to any store, we can approach this problem using binary search. The idea is to perform a binary search on the possible values for the maximum number of products given to any store. For each mid value in the binary search, we check if it is possible to distribute the products in a way that ensures the maximum number of products given to any store is less than or equal to the mid value. If it is possible, we update our result and continue the binary search on the left half; otherwise, we search the right half.

Here is the step-by-step algorithm:

Perform binary search on the possible range of the maximum number of products given to any store (left = 1, right = maximum quantity in the array).
For each mid value, try to distribute the products in a way that ensures no store receives more than mid products.
If distribution is possible, update the result and continue the search on the left half by setting right = mid - 1.
If distribution is not possible, search the right half by setting left = mid + 1.
Return the final result.

Time Complexity:

Binary search takes O(log(max(quantities)) time.
Checking the feasibility of a mid value takes O(n) time.
Therefore, the overall time complexity is O(n log(max(quantities))).

Space Complexity:

The algorithm uses only a constant amount of extra space, resulting in O(1) space complexity.

Solution Code

class Solution {
    public int minimizeTheMaximum(int n, int[] quantities) {
        int left = 1, right = Arrays.stream(quantities).max().getAsInt();
        int result = -1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (isDistributionPossible(quantities, n, mid)) {
                result = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        
        return result;
    }
    
    private boolean isDistributionPossible(int[] quantities, int n, int maxProducts) {
        int stores = 0;
        for (int quantity : quantities) {
            stores += (quantity + maxProducts - 1) / maxProducts;
        }
        return stores <= n;
    }
}

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