Minimized Maximum of Products Distributed to Any Store

• Find the maximum quantity max_q in the quantities array.
• Iterate through possible answers x from 1 to max_q.
• For each x, calculate the total number of stores required.
  • Initialize stores_needed = 0.
  • For each quantity q in quantities:
    • stores_needed += ceil(q / x), which is (q + x - 1) / x.
• If stores_needed is less than or equal to n, then x is a valid solution. Since we are iterating from the smallest possible x, this is the minimum possible maximum. Return x.

The problem asks to find the minimum possible value of x, where x is the maximum number of products given to any store. The smallest possible value for x is 1. The largest possible value for x is the maximum quantity in the quantities array, as in the worst case, a single store might have to take all products of a certain type. We can iterate through each possible value of x from 1 upwards. For each x, we check if it's a feasible maximum. To check if a given x is feasible, we calculate the total number of stores required. For a product type with q items, if each store can hold at most x items, we need ceil(q / x) stores. In integer arithmetic, this is calculated as (q + x - 1) / x. We sum up the required stores for all product types. If this total is less than or equal to the available stores n, then x is a feasible maximum. Since we are checking x in increasing order (1, 2, 3, ...), the first x that we find to be feasible will be the minimum possible value. This approach is simple to understand but inefficient for large inputs.

1class Solution {2    public int minimizedMaximum(int n, int[] quantities) {3        int maxQuantity = 0;4        for (int q : quantities) {5            maxQuantity = Math.max(maxQuantity, q);6        }7 8        for (int x = 1; x <= maxQuantity; x++) {9            if (canDistribute(n, quantities, x)) {10                return x;11            }12        }13        return maxQuantity; // Should not be reached if constraints are met14    }15 16    private boolean canDistribute(int n, int[] quantities, int x) {17        long storesNeeded = 0;18        for (int q : quantities) {19            storesNeeded += (q + x - 1) / x;20        }21        return storesNeeded <= n;22    }23}

• Define the search space for the answer x. The lower bound low is 1, and the upper bound high is the maximum value in quantities.
• Perform a binary search on this range [low, high].
• In each step of the binary search, pick a mid value.
• Check if it's possible to distribute all products with mid as the maximum per store. This is done by a helper function canDistribute(mid).
  • canDistribute(mid) calculates the total stores needed: sum(ceil(q / mid)) for all q in quantities.
  • It returns true if the total stores needed is <= n, false otherwise.
• If canDistribute(mid) is true, it means mid is a possible answer. We try for a smaller answer, so we update ans = mid and set high = mid - 1.
• If canDistribute(mid) is false, mid is too small. We need a larger maximum, so we set low = mid + 1.
• The loop continues until low > high, and the final value of ans is the minimized maximum.

The core idea is to search for the answer x (the minimized maximum) instead of constructing the distribution. The range of possible answers for x is from 1 to max(quantities). We can define a function canDistribute(x) that returns true if it's possible to distribute all products with x as the maximum per store, and false otherwise. The canDistribute(x) function works by calculating the total number of stores required for a given x. For each product quantity q, the number of stores needed is ceil(q / x), which can be calculated as (q + x - 1) / x. We sum these values for all quantities. If the total number of stores needed is not more than n, then canDistribute(x) is true. The property we leverage is: if canDistribute(x) is true, then canDistribute(x+1) is also true. This creates a monotonic sequence of false, false, ..., true, true for the canDistribute function over the range of x, allowing us to use binary search to find the first true value, which corresponds to the minimum x. The binary search algorithm is as follows:

• Set the search range: low = 1, high = max(quantities).
• Initialize a variable ans to store the best possible answer, e.g., ans = high.
• While low <= high:
  • Calculate mid = low + (high - low) / 2.
  • If canDistribute(mid) is true:
    • mid is a potential answer. We try to find an even smaller one.
    • So, we record mid as our current best answer (ans = mid) and search in the lower half (high = mid - 1).
  • Else (canDistribute(mid) is false):
    • mid is too small to be a valid maximum. We need to allow stores to take more items.
    • We search in the upper half (low = mid + 1).
• After the loop, ans will hold the minimized maximum.

1class Solution {2    public int minimizedMaximum(int n, int[] quantities) {3        int low = 1;4        int high = 0;5        for (int q : quantities) {6            high = Math.max(high, q);7        }8 9        int ans = high;10 11        while (low <= high) {12            int mid = low + (high - low) / 2;13            if (mid == 0) { // Avoid division by zero, though low starts at 114                low = 1;15                continue;16            }17            18            if (canDistribute(n, quantities, mid)) {19                ans = mid;20                high = mid - 1;21            } else {22                low = mid + 1;23            }24        }25        return ans;26    }27 28    private boolean canDistribute(int n, int[] quantities, int x) {29        long storesNeeded = 0;30        for (int q : quantities) {31            storesNeeded += (long)(q + x - 1) / x;32        }33        return storesNeeded <= n;34    }35}