Program to find minimum limit of balls in a bag in Python

Suppose we have an array nums where the ith element indicates a bag containing nums[i] number of balls. We also have another value called mx. We can perform the following operation at most mx times −

• Select any bag of balls and divide it into two new bags with at least one ball.

• Here penalty is the maximum number of balls in a bag.

We have to minimize the penalty after the operations. So finally, we have to find the minimum possible penalty after performing the operations.

So, if the input is like nums = [4,8,16,4], mx = 4, then the output will be 4 because we can perform following operations: Initially we have bags like [4,8,16,4], split bag with 16 balls like [4,8,8,8,4], then for each bag with 8 balls, divide them into two bags with 4 balls each, so array will be like [4,4,4,8,8,4], then [4,4,4,4,4,8,4] and finally [4,4,4,4,4,4,4,4], so here minimum we have 4 balls, that is the penalty.

Algorithm Steps

To solve this, we will follow these steps −

• Define a function helper(). This will take target, mx

• if target is same as 0, then

  • return mx + 1

• count := 0

• for each num in nums, do

  • count := count + quotient of (num - 1)/target

• return count <= mx

• From the main method, do the following

• left := maximum of the quotient of (sum of all elements of nums /(size of nums + mx)) and 1

• right := maximum of nums

• while left < right, do

  • mid := quotient of (left + right)/2

  • if helper(mid, mx) is non-zero, then

    • right := mid

  • otherwise,

    • left := mid + 1

• return left

Example

Let us see the following implementation to get better understanding −

def solve(nums, mx):
    def helper(target, mx):
        if target == 0:
            return mx + 1
        count = 0
        for num in nums:
            count += (num - 1) // target
        return count <= mx

    left, right = max(sum(nums) // (len(nums) + mx), 1), max(nums)
    while left < right:
        mid = (left + right) // 2
        if helper(mid, mx):
            right = mid
        else:
            left = mid + 1
    return left

# Test with the given example
nums = [4, 8, 16, 4]
mx = 4
result = solve(nums, mx)
print(f"Minimum penalty: {result}")

Minimum penalty: 4

How It Works

The solution uses binary search to find the minimum penalty. The helper() function checks if a given penalty target is achievable with at most mx operations. For each bag, it calculates how many splits are needed to reduce it to the target size using the formula (num - 1) // target.

The binary search bounds are set between a theoretical minimum (based on distributing all balls evenly) and the maximum bag size. This approach efficiently finds the optimal penalty without trying all possible combinations.

Conclusion

This problem demonstrates an effective use of binary search for optimization. By checking if a penalty is achievable and adjusting the search bounds accordingly, we can find the minimum possible penalty efficiently.