This is a classic optimization problem found in interviews, often referred to as "min-max" optimization. The LeetCode 2064 problem, "Minimized Maximum of Products Distributed to Any Store," tests your ability to recognize when a solution space is monotonic and can be searched efficiently.

Brute Force Approach for Minimized Maximum of Products Distributed to Any Store

A naive approach involves checking every possible value for the maximum load $x$, starting from the smallest possible valid load up to the largest possible load.

The smallest possible load is 1 (assuming at least one product exists). The largest possible load occurs if we dump an entire product type into a single store, which would be max(quantities).

The brute force algorithm works as follows:

1. Start with a candidate capacity $x = 1$.
2. Check if it is possible to distribute all products such that no store receives more than $x$ products and the total stores used is $\le n$.
3. If it is not possible, increment $x$ to 2, then 3, and so on.
4. The first $x$ that satisfies the condition is our answer.

Pseudo-code:

text
max_val = max(quantities)
for x from 1 to max_val:
    stores_needed = 0
    for q in quantities:
        stores_needed += ceiling(q / x)
    
    if stores_needed <= n:
        return x

Complexity Analysis: The check function iterates through quantities, taking $O(m)$ time. In the worst case, the loop runs max(quantities) times.

• Time Complexity: $O(m \cdot \max(quantities))$. Given constraints where quantities can be up to $10^5$, this results in roughly $10^{10}$ operations, leading to a Time Limit Exceeded (TLE) error.
• Space Complexity: $O(1)$.

Core Insight for Minimized Maximum of Products Distributed to Any Store

The core insight lies in inverting the problem: instead of trying to calculate the optimal distribution directly, we guess a value for the maximum capacity and verify if it is valid.

The Condition Function: We need a function canDistribute(k) that returns true if it is possible to distribute all products such that no store gets more than $k$ items. To implement this efficiently:

1. Iterate through each product type in quantities.
2. For a product type with quantity $Q$, calculate how many stores are required if each store holds at most $k$ items. This is calculated as $\lceil Q / k \rceil$.
3. Sum the required stores for all product types.
4. If the total count of required stores is less than or equal to $n$, the capacity $k$ is feasible.

Visualizing the Search: Imagine the range of possible answers from 1 to $100,000$ (the max quantity).

1. We pick the midpoint.
2. We calculate the total stores needed.
3. If the total stores needed is $\le n$, this midpoint is a valid maximum. However, we want to minimize this maximum, so we try to find a smaller valid value in the left half of the range.
4. If the total stores needed is $> n$, the midpoint is too small (it forces us to use too many stores). We must look for a larger value in the right half.

Execution Flow

Consider Example 1: n = 6, quantities = [11, 6].

1. Initialization:

   • left = 1
   • right = 11 (max of [11, 6])

2. Iteration 1:

   • mid = (1 + 11) / 2 = 6.
   • Check(6):
     • Type 0 (11 items): $\lceil 11/6 \rceil = 2$ stores.
     • Type 1 (6 items): $\lceil 6/6 \rceil = 1$ store.
     • Total stores = 3.
     • $3 \le 6$ is True.
   • mid is valid. Can we do better? Set right = 6.

3. Iteration 2:

   • left = 1, right = 6.
   • mid = (1 + 6) / 2 = 3.
   • Check(3):
     • Type 0 (11 items): $\lceil 11/3 \rceil = 4$ stores.
     • Type 1 (6 items): $\lceil 6/3 \rceil = 2$ stores.

4. Iteration 3:

   • left = 1, right = 3.
   • mid = (1 + 3) / 2 = 2.
   • Check(2):
     • Type 0 (11 items): $\lceil 11/2 \rceil = 6$ stores.
     • Type 1 (6 items): $\lceil 6/2 \rceil = 3$ stores.

5. Termination:

   • left is 3, right is 3. Loop ends.
   • Return 3.

Proof of Correctness

The correctness relies on the invariant maintained by the binary search logic:

1. Invariant: The optimal solution always lies within the interval [left, right].
2. Monotonicity: The function stores_needed(k) is non-increasing. As capacity $k$ increases, the number of stores needed decreases or stays the same.
3. Boundary Convergence:
   • If mid is feasible, mid could be the optimal minimum, or the optimal is smaller. By setting right = mid, we ensure we don't discard the potential optimal value.
   • If mid is not feasible, mid is strictly too small. The optimal value must be $> mid$. By setting left = mid + 1, we shrink the search space correctly.
4. Since the range is finite and decreases at every step, the algorithm is guaranteed to terminate at the smallest feasible value.

Reference Implementation

C++ Solution for LeetCode 2064

cpp
class Solution {
public:
    int minimizedMaximum(int n, vector<int>& quantities) {
        // Search space: from 1 to the max quantity
        int left = 1;
        int right = *max_element(quantities.begin(), quantities.end());
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            
            if (canDistribute(n, quantities, mid)) {
                // If feasible, try a smaller maximum (or stick with current)
                right = mid;
            } else {
                // If not feasible, we need a larger maximum capacity
                left = mid + 1;
            }
        }
        
        return left;
    }

private:
    // Helper function to check if distribution is possible with max capacity 'k'
    bool canDistribute(int n, const vector<int>& quantities, int k) {
        int storesNeeded = 0;
        for (int q : quantities) {
            // Calculate ceil(q / k) using integer arithmetic
            storesNeeded += (q + k - 1) / k;
            
            // Optimization: Early exit if we exceed available stores
            if (storesNeeded > n) return false;
        }
        return storesNeeded <= n;
    }
};

Java Solution for LeetCode 2064

java
class Solution {
    public int minimizedMaximum(int n, int[] quantities) {
        int left = 1;
        int right = 0;
        
        // Find the maximum quantity to set the upper bound
        for (int q : quantities) {
            right = Math.max(right, q);
        }
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            
            if (canDistribute(n, quantities, mid)) {
                // Mid is a valid max capacity, try to minimize it further
                right = mid;
            } else {
                // Mid is too small, cannot fit products in n stores
                left = mid + 1;
            }
        }
        
        return left;
    }
    
    private boolean canDistribute(int n, int[] quantities, int k) {
        int storesNeeded = 0;
        for (int q : quantities) {
            // Equivalent to Math.ceil((double)q / k)
            storesNeeded += (q + k - 1) / k;
            
            if (storesNeeded > n) return false;
        }
        return true;
    }
}

Python Solution for LeetCode 2064

python
import math

class Solution:
    def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
        left, right = 1, max(quantities)
        
        while left < right:
            mid = (left + right) // 2
            
            # Check feasibility for capacity 'mid'
            if self.can_distribute(n, quantities, mid):
                right = mid
            else:
                left = mid + 1
                
        return left

    def can_distribute(self, n: int, quantities: List[int], k: int) -> bool:
        stores_needed = 0
        for q in quantities:
            # Calculate ceiling of q / k
            stores_needed += math.ceil(q / k)
            if stores_needed > n:
                return False
        return True

In all these cases, we define a search space for the answer and write a greedy helper function to validate a specific candidate value.