Minimized Maximum of Products Distributed to Any Store

class Solution {
public:
    int minimizedMaximum(int n, vector<int>& quantities) {
        int m = quantities.size();

        auto comparator = [](pair<int, int>& x, pair<int, int>& y) {
            return (long long)x.first * y.second < (long long)x.second * y.first;
        };

        vector<pair<int, int>> products;

        for (int i = 0; i < m; i++) {
            products.push_back({quantities[i], 1});
        }

        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(comparator)>
            pq(products.begin(), products.begin() + m);

        for (int i = 0; i < n - m; i++) {
            pair<int, int> maxProduct = pq.top();
            int quantity = maxProduct.first;
            int stores = maxProduct.second;
            pq.pop();
            pq.push({quantity, stores + 1});
        }

        pair<int, int> maxProduct = pq.top();
        int quantity = maxProduct.first;
        int stores = maxProduct.second;

        return ceil((double)quantity / stores);
    }
};

In the given C++ code, the primary goal is to determine the minimized maximum number of products a store must carry when distributing given quantities of products to a specified number of stores while ensuring the distribution is as even as possible.

• The function minimizedMaximum accepts two parameters, n - the number of stores, and quantities - a vector container representing the quantity of different products.
• The use of a priority queue becomes evident as it helps sort the products based on the ratio of their quantity to the number of stores they have been assigned, ensuring the distribution is as balanced as possible.
• The priority queue is initialized using product quantities coupled with a store count of 1 for each product, ensuring each product is initially considered for one store allocation.

The distribution algorithm functions as follows:

1. Start by pushing each of the elements of quantities along with an initial assignment to one store into a vector products, creating pairs. These pairs keep track of the quantity of the product and how many stores it is assigned to.
2. Initialize a priority queue (with properties and ordering defined by the comparator lambda function). This queue will keep the products with the highest quantity to the number of assigned stores ratio at the top, ensuring that the 'heaviest' loads are always considered for further distribution into another store.
3. For each product, if there are stores left which do not have any product assigned (n - m stores in case there are more stores than products), extract the top product from the queue (i.e., the one with the highest load per store), increment its store count, and push it back to the queue adjusted for the increased store count.
4. Once all products are fully distributed across all stores, extract the top element from the queue again. This pair now contains the product which has the highest count per store out of all possible distributions.
5. Calculate the output by dividing the product quantity by the number of stores assigned to it and rounding up. This gives the minimized maximum number of products any store will carry.
6. Return this value as the result.

The function employs a mixture of greedy and sorting techniques facilitated by a priority queue to ensure that the product quantities are as evenly distributed as possible across all available stores.

class Solution {

    public int calculateMinimizedMaximum(int totalStores, int[] productQuantities) {
        int productCount = productQuantities.length;

        // Pair list creation for managing stores and quantities
        List<int[]> storeProductMapping = new ArrayList<>();

        // Assignment of one store to each product initially
        for (int i = 0; i < productCount; i++) {
            storeProductMapping.add(new int[] { productQuantities[i], 1 });
        }

        // Priority Queue to maintain the distribution of stores
        PriorityQueue<int[]> storeProductQueue = new PriorityQueue<>((a, b) ->
            Long.compare((long) b[0] * a[1], (long) a[0] * b[1])
        );

        // Populating the priority queue
        storeProductQueue.addAll(storeProductMapping);

        // Distribute the remaining stores
        for (int i = 0; i < totalStores - productCount; i++) {
            // Retrieve the product type with the current critical ratio
            int[] currentMaxProduct = storeProductQueue.poll();
            int quantity = currentMaxProduct[0];
            int assignedStores = currentMaxProduct[1];

            // Update the queue
            storeProductQueue.offer(
                new int[] { quantity, assignedStores + 1 }
            );
        }

        // Fetch the product type that will have the max value after distribution
        int[] finalMaxProduct = storeProductQueue.poll();
        int finalQuantity = finalMaxProduct[0];
        int finalAssignedStores = finalMaxProduct[1];

        // Calculate the minimized maximum ratio
        return (int) Math.ceil(
            (double) finalQuantity / finalAssignedStores
        );
    }
}

This Java program addresses the problem of finding the minimized maximum number of products that a store receives based on a distribution of a specific number of stores and given product quantities. The solution adopted here involves several key steps:

• Create a priority queue to manage the distribution such that the most imbalanced ratio of product quantity to stores is adjusted first. This is crucial to ensuring that at each step, the distribution remains as even as possible across all stores.
• Initially assign each product to at least one store, and populate the priority queue with these initial allocations. Each entry in the queue is an array where the first element is the quantity of the product, and the second element is the number of stores it is assigned to.
• Distribute the remaining stores one by one to the product which, at that moment, has the highest ratio of quantity to the number of stores. This ensures that no single store ends up with a disproportionately high number of products.
• After all stores have been assigned, calculate the minimized maximum number of products per store by examining the top of the priority queue.

The algorithm effectively spreads out product quantities across stores in such a way that the highest quantity any store has to manage is as low as possible. This method optimizes store product capacity utilization and can be particularly useful in resource distribution and supply chain optimization scenarios.

class Solution:
    def findMinimizedMax(self, num_stores, product_quantities):
        count_products = len(product_quantities)

        # Creating tuple array for heap logic
        product_heap = [(-quantity / 1, quantity, 1) for quantity in product_quantities]

        # Convert array to a heap
        heapq.heapify(product_heap)

        # Distribute remaining stores
        for _ in range(num_stores - count_products):
            # Extract tuple with highest ratio
            (
                current_neg_ratio,
                current_quantity,
                current_stores_assigned,
            ) = heapq.heappop(product_heap)

            # Assign one more store and calculate updated max need
            new_stores_assigned = current_stores_assigned + 1
            new_neg_ratio = -current_quantity / new_stores_assigned

            # Push updated values into heap
            heapq.heappush(
                product_heap,
                (
                    new_neg_ratio,
                    current_quantity,
                    new_stores_assigned,
                ),
            )

        # Get the top element of the heap
        _, max_product_quantity, final_store_count = heapq.heappop(product_heap)

        # Calculate the required maximum number of products per store
        return math.ceil(max_product_quantity / final_store_count)

To solve the problem of determining the minimized maximum of products distributed to any store, you employ Python 3 with the use of heaps for optimal efficiency. Start by converting your list of product quantities into a heap where each element is a tuple representing the negative ratio of quantity to stores, the quantity itself, and the number of stores assigned. This is done to maximize the efficiency of retrieving the product with the currently highest needed capacity per store.

• Initialize your heap with each product being distributed to one store initially.
• For each additional store yet to be assigned, perform the following:
  1. Extract the product with the highest demand per store.
  2. Reassign one additional store to this product, recalculate the ratio, and insert the new values back into the heap.
• After all stores are distributed, the tuple on the top of the heap will have the highest product quantity to store ratio.
• Calculate the answer by taking the ceiling of the ratio of product quantity to stores in this tuple.

This method ensures that the maximum quantity of any product a store has to handle is minimized, considering the distribution constraints of the total number of stores and product quantities. Use the heapq library for heap operations and math.ceil for rounding off calculations.