What is the brute force solution for Minimum Number of Groups to Create a Valid Assignment?

The brute force approach for Minimum Number of Groups to Create a Valid Assignment is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will try to generate all possible distributions of balls into boxes and check which distributions meet the criteria of having a maximum size difference of one between the largest and smallest box. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Groups to Create a Valid Assignment?

Minimum Number of Groups to Create a Valid Assignment uses the following concepts: Array, Hash Table, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Groups to Create a Valid Assignment?

Always start by understanding the constraints and requirements. Think about edge cases, such as when all balls are the same. Practice breaking down the problem into smaller parts.

What are common mistakes when solving Minimum Number of Groups to Create a Valid Assignment?

Not calculating frequencies correctly. Overcomplicating the problem by trying to generate all combinations.

#2910

Minimum Number of Groups to Create a Valid Assignment

Medium

ArrayHash TableGreedyHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

In this approach, we calculate the frequency of each ball and determine the minimum number of boxes required based on the maximum frequency of any ball. This allows us to efficiently distribute the balls while adhering to the size difference constraint.

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Algorithm

3 steps

1Step 1: Calculate the frequency of each ball number.
2Step 2: Determine the maximum frequency from the frequency map.
3Step 3: The minimum number of boxes needed is equal to the maximum frequency, as each box can hold one ball of the maximum frequency, and the rest can be distributed accordingly.

solution.py6 lines

1from collections import Counter
2
3def min_boxes_opt(balls):
4    freq = Counter(balls)
5    max_freq = max(freq.values())
6    return max_freq

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Complexity note: This complexity is efficient because we only traverse the list of balls once to count frequencies and once more to find the maximum frequency, making it linear in relation to the input size.

1The maximum frequency of any ball dictates the minimum number of boxes needed.
2Distributing balls evenly while adhering to the size difference constraint is crucial.

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