What is the time complexity of Minimized Maximum of Products Distributed to Any Store?

The optimal solution for Minimized Maximum of Products Distributed to Any Store runs in O(n log m) time and uses O(1) space. This complexity arises because we perform a binary search (log m) on the maximum quantity and for each mid, we check distribution (O(n)).

What is the brute force solution for Minimized Maximum of Products Distributed to Any Store?

The brute force approach for Minimized Maximum of Products Distributed to Any Store is Brute Force, with O(n * m) time complexity and O(1) space complexity. The brute force approach involves trying every possible maximum number of products per store and checking if we can distribute all products under that limit. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Minimized Maximum of Products Distributed to Any Store?

Minimized Maximum of Products Distributed to Any Store uses the following concepts: Array, Binary Search, Greedy. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Minimized Maximum of Products Distributed to Any Store?

Always consider if binary search can be applied in optimization problems. Clearly define helper functions to check conditions (like distribution) to keep your code clean. Practice explaining your thought process as you code; it helps clarify your understanding.

What are common mistakes when solving Minimized Maximum of Products Distributed to Any Store?

Failing to recognize the binary search nature of the problem, leading to inefficient solutions. Not properly checking the conditions for valid distribution, which can lead to incorrect results.

#2064

Minimized Maximum of Products Distributed to Any Store

Medium

ArrayBinary SearchGreedyBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

The optimal approach uses binary search to efficiently find the minimum possible maximum number of products per store. This leverages the monotonic nature of the problem, where if a certain x is valid, all larger x are also valid.

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Algorithm

4 steps

1Step 1: Set low to 1 and high to the maximum quantity in the quantities array.
2Step 2: Perform binary search: while low is less than or equal to high, calculate mid as the average of low and high.
3Step 3: Check if mid is a valid maximum by calculating how many stores can be filled with mid products each. If valid, move high to mid - 1; otherwise, move low to mid + 1.
4Step 4: The lowest valid mid found will be the answer.

solution.py14 lines

1def minimizedMaximum(n, quantities):
2    def canDistribute(x):
3        stores_filled = 0
4        for quantity in quantities:
5            stores_filled += quantity // x
6        return stores_filled >= n
7    low, high = 1, max(quantities)
8    while low <= high:
9        mid = (low + high) // 2
10        if canDistribute(mid):
11            high = mid - 1
12        else:
13            low = mid + 1
14    return low

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Complexity note: This complexity arises because we perform a binary search (log m) on the maximum quantity and for each mid, we check distribution (O(n)).

1The problem has a monotonic property: if a certain maximum x is valid, any larger x is also valid.
2Binary search is a powerful technique for optimization problems where you need to minimize or maximize a value.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.