2064. Minimized Maximum of Products Distributed to Any Store - Detailed Explanation

Problem Statement

You are given an integer n representing the number of stores and an integer array quantities where each quantities[i] is the number of units of the i-th product type that you have. You need to distribute all these product units to the stores under the following rule:

• Rule: For each product type, a single store can receive at most one unit (i.e. units of the same product must be distributed to different stores).

Your goal is to minimize the maximum number of products assigned to any single store. In other words, if after the distribution the store with the highest load receives x products, you want to make x as small as possible. Return the minimum possible value of x.

Example 1

• Input:
  n = 5, quantities = [11, 6]
• Output: 4
• Explanation:

  • For product type 0 (11 units): To keep a store’s load at most 4, we need to distribute these 11 units into ⎡11 / 4⎤ = 3 stores.

  • For product type 1 (6 units): We need ⎡6 / 4⎤ = 2 stores.

  • In total, 3 + 2 = 5 stores are required, which is exactly n.
    It can be shown that no smaller maximum value per store is possible.

Example 2

• Input:
  n = 6, quantities = [10, 4, 4]
• Output: 4
• Explanation:

  • For product type 0: ⎡10 / 4⎤ = 3 stores.

  • For product type 1: ⎡4 / 4⎤ = 1 store.

  • For product type 2: ⎡4 / 4⎤ = 1 store.

  • Total required stores = 3 + 1 + 1 = 5 ≤ 6.
    Testing a candidate maximum less than 4 would require more than 6 stores, so the answer is 4.

Example 3

• Input:
  n = 10, quantities = [1, 2, 3, 4, 5]
• Output: 2
• Explanation:
  • For each product type, the number of required stores if we set the maximum allowed per store to 2 are:
    Product 0: ⎡1/2⎤ = 1, Product 1: ⎡2/2⎤ = 1, Product 2: ⎡3/2⎤ = 2, Product 3: ⎡4/2⎤ = 2, Product 4: ⎡5/2⎤ = 3.
    Total = 1 + 1 + 2 + 2 + 3 = 9, which is ≤ 10.
  • Trying any value smaller than 2 would force at least one product type to require more than 10 stores. Hence, the answer is 2.

Constraints

• 1 ≤ quantities.length ≤ 10⁵
• 1 ≤ quantities[i] ≤ 10⁵
• quantities.length ≤ n ≤ 10⁹

2. Hints to Guide Your Approach

1. Observation on Distribution:
   Since a store can receive at most one unit of a particular product type, for each product with quantity q, if you want to limit the number of units per store to a candidate maximum x, you will need at least ⎡q / x⎤ different stores to distribute that product.

2. Checking Feasibility:
   If you sum the number of stores needed for all product types (i.e. for each quantity q, compute ⎡q / x⎤) and if the sum is less than or equal to n, then it is possible to achieve a maximum of x products per store. This observation naturally suggests using binary search on the candidate maximum value x.

Approaches to Solve the Problem

Approach 1: Brute Force (Conceptual)

Explanation

• Idea:
  One could try every possible maximum load per store from 1 up to max(quantities) and check whether it’s possible to distribute the products without exceeding that load.

• Key Steps:

  1. For each candidate maximum x, iterate over each product type.

  2. Calculate the required number of stores for that product as ⎡q / x⎤.

  3. Sum these values and see if they are ≤ n.

• Drawbacks:
  This approach is conceptually simple but would be too slow because x can range up to 10⁵ and there can be up to 10⁵ product types.

Approach 2: Binary Search

Explanation

• Idea:
  Use binary search to efficiently find the smallest x for which the distribution is possible.

• Key Steps:

  1. Define the Search Space:
     The minimum possible maximum load is 1 and the maximum possible load is max(quantities) (if one product type is given entirely to different stores).

  2. Binary Search Loop:

     • For a candidate x, calculate the total number of stores required by summing ⎡q / x⎤ for each product type.

     • If the sum is ≤ n, then x is feasible (try a smaller x to see if you can lower the maximum).

     • Otherwise, x is too small, so search in the higher half.

  3. Return the Answer:
     The first candidate x that is feasible will be the minimized maximum number of products per store.

Visual Example (Example 1)

For quantities = [11, 6] and n = 5:

• Search Space: x in [1, 11]

• Candidate x = 6:

  • Product 0: ⎡11/6⎤ = 2, Product 1: ⎡6/6⎤ = 1 → Total = 3 (feasible, but can we do lower?)

• Candidate x = 3:

  • Product 0: ⎡11/3⎤ = 4, Product 1: ⎡6/3⎤ = 2 → Total = 6 (not feasible)

• Narrow Down:
  Through binary search you will eventually determine that x = 4 is the smallest value for which the total required stores is exactly 5.

Complexity Analysis

• Time Complexity:
  Each binary search iteration processes all product types (O(m) where m = number of product types) and there are O(log(max(quantities))) iterations.

  Total: O(m · log(max(quantities))).

• Space Complexity: O(1)

Code Implementations

Python Code

Below is the Python implementation using the binary search approach: