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Minimum Limit of Balls in a Bag

Difficulty: Medium

Problem Description

You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations. You can perform the following operation at most maxOperations times: Take any bag of balls and divide it into two new bags with a positive number of balls. Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations. Return the minimum possible penalty after performing the operations.

Key Insights

  • The goal is to minimize the maximum number of balls in any bag after performing a limited number of operations.
  • A binary search approach can be applied to efficiently find the minimum possible penalty.
  • For each potential penalty value during the binary search, a helper function can determine if that penalty can be achieved within the allowed number of operations.
  • The division of bags needs to be optimized; larger bags require more operations to reduce their size effectively.

Space and Time Complexity

Time Complexity: O(n log(max(nums))) - where n is the number of bags and max(nums) is the maximum number of balls in any bag. The log(max(nums)) factor arises from the binary search over the possible penalties.

Space Complexity: O(1) - only a few variables are used for calculations, regardless of input size.

Solution

To solve the problem, we will use binary search on the possible maximum penalty values. We will define the range for the penalty from 1 to the maximum number of balls in any bag. For each mid-point in our binary search, we will check if it is possible to achieve that penalty by simulating the division of bags. We use a counter to track the number of operations needed to achieve the desired penalty.

Code Solutions

def minimumLimit(nums, maxOperations):
    def canAchievePenalty(penalty):
        operations = 0
        for balls in nums:
            if balls > penalty:
                operations += (balls - 1) // penalty
        return operations <= maxOperations

    left, right = 1, max(nums)
    while left < right:
        mid = (left + right) // 2
        if canAchievePenalty(mid):
            right = mid
        else:
            left = mid + 1
    return left
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