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Minimum Number of Groups to Create a Valid Assignment

Difficulty: Medium

Problem Description

You are given a collection of numbered balls and instructed to sort them into boxes for a nearly balanced distribution. There are two rules you must follow:

  1. Balls with the same box must have the same value. But, if you have more than one ball with the same number, you can put them in different boxes.
  2. The biggest box can only have one more ball than the smallest box.

Return the fewest number of boxes to sort these balls following these rules.

Key Insights

  • Each unique ball number can be distributed across different boxes.
  • The distribution must ensure the size difference between the largest and smallest box is at most one.
  • The maximum frequency of any ball number will dictate how many boxes are needed, as it will need to be evenly distributed to comply with the size constraints.

Space and Time Complexity

Time Complexity: O(n), where n is the number of balls (length of the input array). Space Complexity: O(m), where m is the number of unique ball numbers.

Solution

To solve the problem, we can use a hash table (or dictionary) to count the frequency of each ball number. The maximum frequency among the ball numbers will determine the number of boxes needed. By distributing the balls evenly, we ensure that the size difference between the boxes remains within the allowed limit.

  1. Count the frequency of each ball number using a hash table.
  2. Determine the maximum frequency from the hash table.
  3. Calculate the number of boxes required by distributing the balls based on the maximum frequency.

Code Solutions

def minBoxes(balls):
    from collections import Counter
    
    # Count the frequency of each ball number
    frequency = Counter(balls)
    
    # Get the maximum frequency
    max_freq = max(frequency.values())
    
    # The number of boxes needed is equal to the maximum frequency
    return max_freq

# Example cases
print(minBoxes([3,2,3,2,3]))  # Output: 2
print(minBoxes([10,10,10,3,1,1]))  # Output: 4
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