Solution Explanation: Minimum Limit of Balls in a Bag

This problem asks to find the minimum possible maximum number of balls in any bag after performing at most maxOperations divisions. The core idea is to use binary search to efficiently find this minimum value.

Intuition:

The problem inherently has a monotonically decreasing relationship between the maximum number of balls in a bag and the number of operations needed. If we allow larger maximum bag sizes, we need fewer operations. Conversely, smaller maximum bag sizes require more operations. This monotonicity makes binary search an ideal approach.

Algorithm:

1. Binary Search Space: The search space for the minimum penalty (maximum balls in a bag) is from 1 (minimum possible value) to the maximum value in nums (maximum possible value).

2. Check Function: We create a helper function check(mid) which determines if a given mid (potential maximum number of balls) is achievable within the given maxOperations. This function iterates through nums, calculating the number of operations needed to reduce each bag size to at most mid. If the total operations are within the limit, it returns true; otherwise, it returns false. The number of operations needed for a bag of size x is (x - 1) // mid. Integer division (// or / depending on the language) is used to find the number of splits needed.

3. Binary Search: The main part of the algorithm uses binary search. It starts with left = 1 and right = max(nums). In each iteration:

   • It calculates the midpoint mid = (left + right) // 2.
   • It calls check(mid) to see if mid is a valid solution.
   • If check(mid) is true, it means we can achieve a maximum bag size of mid within the operation limit. Thus, we update right = mid to explore even smaller maximum bag sizes.
   • If check(mid) is false, we need a larger maximum bag size, so we update left = mid + 1.

4. Result: The loop continues until left and right converge. At this point, left holds the minimum possible maximum number of balls in a bag.

Time Complexity: O(N log M), where N is the length of nums and M is the maximum value in nums. The binary search takes O(log M) iterations, and each iteration involves iterating through nums (O(N)) to check the feasibility of a potential solution.

Space Complexity: O(1). The algorithm uses only a constant amount of extra space.

Code Examples (Python and Java):

Python:

import bisect
 
class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        def check(x):
            return sum((num - 1) // x for num in nums) <= maxOperations
 
        left, right = 1, max(nums)
        return bisect_left(range(1, right + 1), True, key=check)

Java:

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int left = 1, right = Arrays.stream(nums).max().getAsInt();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (check(nums, mid, maxOperations)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
 
    private boolean check(int[] nums, int mid, int maxOperations) {
        long operations = 0;
        for (int num : nums) {
            operations += (num - 1) / mid;
        }
        return operations <= maxOperations;
    }
}

The other language examples provided earlier follow the same basic structure, adapting the syntax and libraries as needed for each language. The core binary search and check function remain consistent across all implementations.