#2910

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Number of Groups to Create a Valid Assignment

The problem involves sorting balls into boxes while minimizing the number of boxes, adhering to size constraints.

Problem Statement

You are given a list of balls, each labeled with a number. You need to sort these balls into boxes while following two rules: each box should have a nearly equal number of balls, and no box should contain more than one ball of the same number. Return the fewest number of boxes that can accommodate all the balls while following these rules.

The problem requires calculating the frequency of each ball number to determine how many boxes are necessary to balance the distribution. A greedy approach, combined with array scanning and hash lookups, helps in finding the optimal solution.

Examples

Example 1

Input: balls = [3,2,3,2,3]

Output: 2

We can sort balls into boxes as follows: The size difference between the two boxes doesn't exceed one.

Example 2

Input: balls = [10,10,10,3,1,1]

Output: 4

We can sort balls into boxes as follows: You can't use fewer than four boxes while still following the rules. For example, putting all three balls numbered 10 in one box would break the rule about the maximum size difference between boxes.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109

Solution Approach

Frequency Calculation

The first step is to calculate the frequency of each number in the input array. This can be done using a hash map or hash table, where the keys are the ball numbers and the values are their respective counts.

Determine Box Count

After calculating the frequencies, the number of boxes required is determined by the highest frequency of any ball. This is because the most frequent ball will dictate the minimum number of boxes needed to maintain a balanced distribution.

Greedy Distribution

Finally, distribute the balls using a greedy approach, filling the boxes while ensuring the size difference between any two boxes doesn’t exceed one. This minimizes the number of boxes while adhering to the constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen for calculating frequencies and distributing balls. If a hash map is used, the time complexity is O(n) where n is the number of balls. Space complexity depends on the number of distinct ball numbers, which could also be O(n) in the worst case.

What Interviewers Usually Probe

Ensure the candidate correctly identifies the use of a hash map for frequency calculation.
Look for understanding of greedy algorithms and the significance of the most frequent number.
Evaluate how the candidate balances the box distribution to avoid exceeding the maximum allowed size difference.

Common Pitfalls or Variants

Common pitfalls

Candidates may fail to consider edge cases where the ball numbers are very skewed in frequency.
Misunderstanding the problem can lead candidates to overcomplicate the solution or underutilize hash maps.
Failing to account for the greedy approach could lead to inefficient solutions or incorrect box counts.

Follow-up variants

What if the balls are not sorted?
How does the solution change if balls must be grouped into fixed-size boxes?
How would the approach differ if the maximum allowed box size was reduced?

FAQ

What is the time complexity of this problem?

The time complexity is O(n) for counting frequencies using a hash map, where n is the number of balls.

What should I consider when determining the number of boxes?

The number of boxes is determined by the highest frequency of any ball. Ensure that the distribution between boxes remains as balanced as possible.

What is the greedy approach used in this problem?

The greedy approach ensures that the distribution of balls into boxes minimizes the number of boxes by filling boxes to balance their sizes while adhering to the constraints.

How does the frequency of balls impact the solution?

The frequency of balls dictates the minimum number of boxes required, as the most frequent ball determines the least number of boxes needed to avoid exceeding the maximum size difference.

Can the approach change if there are fewer boxes than balls?

Yes, in that case, you would need to adjust your strategy, possibly considering constraints on the number of available boxes or adapting to a different problem variant.

terminal

Solution

Solution 1: Hash Table + Enumeration

We use a hash table $cnt$ to count the number of occurrences of each number in the array $nums$. Let $k$ be the minimum value of the number of occurrences, and then we can enumerate the size of the groups in the range $[k,..1]$. Since the difference in size between each group is not more than $1$, the group size can be either $k$ or $k+1$.

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class Solution:
    def minGroupsForValidAssignment(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        for k in range(min(cnt.values()), 0, -1):
            ans = 0
            for v in cnt.values():
                if v // k < v % k:
                    ans = 0
                    break
                ans += (v + k) // (k + 1)
            if ans:
                return ans

Continue Practicing

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