Minimum Limit of Balls in a Bag

class Solution {
public:
    int minSize(vector<int>& arr, int maxOps) {
        int minVal = 1, maxVal = 0;

        for (int value : arr) {
            maxVal = max(maxVal, value);
        }

        while (minVal < maxVal) {
            int mid = (minVal + maxVal) / 2;

            if (canDistribute(mid, arr, maxOps)) {
                maxVal = mid;
            } else {
                minVal = mid + 1;
            }
        }

        return minVal;
    }

private:
    bool canDistribute(int maxBalls, vector<int>& arr, int maxOps) {
        int usedOps = 0;

        for (int number : arr) {
            int neededOps = (number + maxBalls - 1) / maxBalls - 1;
            usedOps += neededOps;

            if (usedOps > maxOps) return false;
        }

        return true;
    }
};

The problem "Minimum Limit of Balls in a Bag" asks you to determine the smallest possible maximum number of balls in any single bag, given that you can make a certain number of operations. Each operation consists of dividing the contents of a bag into two smaller bags.

The solution approaches the problem using a binary search algorithm, implemented in C++. The binary search is performed on possible maximum values of the number of balls per bag. Here's the step-by-step process encapsulated in the given C++ solution:

1. Identify the possible range for the maximum number of balls in a bag, where minVal is initialized to 1 and maxVal finds the maximum number of balls present in any single bag of the input array arr.

2. Implement a binary search between minVal and maxVal. During each iteration, compute the middle value mid and check if it's feasible to ensure that no bag has more than mid balls using no more than the allowed number of operations.

3. Implement a helper function canDistribute, which checks if it's possible to limit the maximum number of balls per bag to maxBalls using the current mid value. It calculates the operations needed for each bag and aggregates these. If at any point the total operations exceed maxOps, it concludes that maxBalls is too low.

4. If canDistribute returns true, adjust maxVal to mid to try a smaller maximum; if false, adjust minVal to mid + 1 to try a larger maximum.

5. Once the binary search concludes, minVal will hold the smallest possible maximum number of balls that can be achieved within the allowed operations.

The solution is efficient, managing the complexity with a logarithmic search pattern in combination with a linear check through all bags in the canDistribute function. This method offers a practical way to handle potentially large input sizes effectively.

class Solution {

    public int smallestMaxBagSize(int[] bags, int maxOps) {
        int min = 1;
        int max = 0;

        for (int bag : bags) {
            max = Math.max(max, bag);
        }

        while (min < max) {
            int mid = (min + max) / 2;
            if (canDistribute(mid, bags, maxOps)) {
                max = mid;
            } else {
                min = mid + 1;
            }
        }

        return min;
    }

    private boolean canDistribute(
        int maxCapacity,
        int[] bags,
        int allowedOps
    ) {
        int usedOps = 0;

        for (int bag : bags) {
            int splits = (int) Math.ceil(bag / (double) maxCapacity) - 1;
            usedOps += splits;

            if (usedOps > allowedOps) {
                return false;
            }
        }

        return true;
    }
}

This Java solution addresses the problem of finding the minimum maximum number of balls that should be present in any bag, adjusting other bags such that the number of operations used to redistribute the balls does not exceed a given maximum (maxOps). The principle method used to solve this problem is binary search, aiming to minimize the maximum ball count in the most filled bag after the allowed redistributions.

• Start by identifying the range of possible maximum numbers of balls in a bag by setting min to 1 (since a bag must have at least one ball) and max to the highest count of balls present in any single bag from the input array.
• Use binary search within this min to max range to efficiently find the smallest possible maximum number of balls per bag.
  • Calculate the midpoint (mid) for the current range and use it as a candidate maximum ball count per bag.
  • A helper function canDistribute is used to check whether it's possible to reduce the ball count in each bag to at most mid balls using at most maxOps redistributive operations.
    • This function computes the necessary operations for each bag by determining how many times you would need to split the contents to ensure no bag exceeds mid balls.
    • If the total used operations exceed maxOps, the function returns false, indicating the need to increase the mid value.
  • Depending on the result from canDistribute, adjust the min and max values for the binary search range: if redistribution is possible, decrease max (tighten the upper bound); otherwise, increase min (raise the lower bound).
• Once min and max converge, min will hold the smallest maximum number of balls that can be achieved through at most maxOps operations.

The complete method ensures efficiency in finding the desired maximum bag size by narrowing down the range continually using the binary search strategy combined with logical redistribution checks through the auxiliary function.

class Solution:
    def minBagSize(self, numbers, maxOps):
        min_size = 1
        max_size = max(numbers)

        while min_size < max_size:
            mid = (min_size + max_size) // 2
            if self.isValid(mid, numbers, maxOps):
                max_size = mid
            else:
                min_size = mid + 1

        return min_size

    def isValid(self, target, numbers, maxOps):
        operations_needed = 0
        for number in numbers:
            ops = (number - 1) // target
            operations_needed += ops
            if operations_needed > maxOps:
                return False
        return True

The Python solution provided addresses the problem of determining the minimum possible size that a ball can be divided into so that all bags contain a maximum number of balls not exceeding the given limit using a specified maximum number of operations.

To achieve this, the solution uses a binary search technique:

• It initializes two boundary variables, min_size and max_size, setting them to 1 and the maximum number found in the numbers list respectively.
• Within a loop, the solution adjusts these boundaries to find the optimal minimum size:
  • It calculates the mid-point and checks if it's a valid solution using the isValid function.
  • If valid, the upper boundary max_size is updated to mid, tightening the search space.
  • If not valid, the lower boundary min_size is increased to mid + 1, narrowing the search scope.
• The process loops until the smallest possible size (min_size) which meets the criteria is found.

The isValid helper function assesses if it's possible to limit the number of balls per bag to the target value without exceeding the maxOps allowed operations:

• It computes the operations needed for each number in the list to ensure no more balls than the target number remain in any bag.
• The operation count for each number is the integer division result of (number - 1) // target, and validations are performed repeatedly across the list of numbers.
• If the total operations required exceed maxOps, it returns False; otherwise, after the loop, it returns True.

By the end of this explanation, you have a clear understanding of how the solution efficiently finds the minimum size of bags using binary search and validation through operations count checks.