We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Minimum Limit of Balls in a Bag

Number: 1886

Difficulty: Medium

Paid? No

Companies: Google, Bloomberg, Intuit, Amazon, Flipkart

Problem Description

Given an array where the i-th element indicates the number of balls in a bag and a limit on the number of operations allowed, you can perform operations by splitting a bag of balls into two positive integer parts. The goal is to minimize the penalty, defined as the maximum number of balls in any bag, by optimally splitting bags up to a given operation limit.

Key Insights

  • The penalty is the maximum bag size after splitting the balls.
  • You can transform larger bags into smaller ones by splitting.
  • For any candidate penalty, you can compute the required number of splits by summing up for each bag as (num - 1) // candidate.
  • Use binary search over possible penalties (from 1 to max(nums)) to identify the minimum feasible penalty with given operations.
  • The check function is crucial: it determines if a given target maximum is achievable with at most maxOperations.

Space and Time Complexity

Time Complexity: O(n log(max(nums))) where n is the number of bags. Space Complexity: O(1) additional space (excluding the input).

Solution

We can decide on a binary search approach on the potential penalties from 1 (best case) to max(nums) (worst case). For each candidate penalty value, calculate the number of operations needed by splitting each bag into parts such that each part has balls count less than or equal to the candidate penalty. The number of splits required for a bag with num balls is (num - 1) // candidate. Sum all operations across bags. If the total is within the allowed maxOperations, then the candidate penalty is achievable and we attempt a smaller penalty; otherwise, we increase the penalty. This binary search efficiently finds the smallest penalty.

Code Solutions

# Python Solution

def minimumSize(nums, maxOperations):
    # Helper function to decide if a given max_ball is feasible by checking required operations.
    def canAchieve(max_ball):
        ops = 0
        for balls in nums:
            # Calculate number of operations needed for current bag.
            ops += (balls - 1) // max_ball
            if ops > maxOperations:  # Early exit if exceeds allowed operations.
                return False
        return True

    # Initialize binary search bounds.
    low, high = 1, max(nums)
    while low < high:
        mid = (low + high) // 2  # Candidate penalty.
        if canAchieve(mid):
            high = mid  # Candidate is feasible, try for a lower penalty.
        else:
            low = mid + 1  # Not feasible, need a larger penalty.
    return low

# Example usage:
print(minimumSize([9], 2))  # Expected output: 3
print(minimumSize([2,4,8,2], 4))  # Expected output: 2
← Back to All Questions