Minimum Limit of Balls in a Bag

• Create a max-heap to store objects or tuples representing {current_penalty, original_size, num_pieces}.
• For each num in the input array nums, add an entry {num, num, 1} to the heap.
• Iterate maxOperations times:
  • Pop the top element {penalty, original_size, num_pieces} from the heap, which represents the current largest penalty contribution.
  • Increment num_pieces by 1, signifying one operation is used to split this bag further.
  • Calculate the new penalty for this bag after the split: new_penalty = ceil(original_size / num_pieces).
  • Push the new state {new_penalty, original_size, num_pieces} back into the heap.
• After the loop, the penalty of the top element in the heap is the minimized maximum penalty.

We can model the problem as follows: for each original bag, we can spend operations to split it into more and more pieces. If an original bag of size s is split into k pieces, it takes k-1 operations, and the new maximum size of these pieces is ceil(s/k). The greedy choice is to always spend one operation on the set of pieces that currently has the highest value.

We use a max-heap to keep track of the current penalty for each original bag. The heap will store tuples or objects containing (current_penalty, original_size, num_pieces). The heap is ordered by current_penalty.

The algorithm is as follows:

1. Initialize a max-heap. For each number num in the input array nums, push a tuple (num, num, 1) into the heap. This represents that initially, each bag is 1 piece of its original size.
2. Loop for maxOperations times: a. Extract the element with the maximum penalty from the heap. Let this be (penalty, original_size, num_pieces). b. We "use" one operation on this original bag. Increment the number of pieces: num_pieces_new = num_pieces + 1. c. Calculate the new penalty for this bag: penalty_new = ceil(original_size / num_pieces_new). d. Insert the new state (penalty_new, original_size, num_pieces_new) back into the heap.
3. After the loop finishes, the top element of the heap contains the maximum penalty in the first element of its tuple. This is the minimum possible penalty.

The ceil(a/b) operation can be calculated using integer arithmetic as (a + b - 1) / b.

1import java.util.PriorityQueue;2 3class Solution {4    public int minimumSize(int[] nums, int maxOperations) {5        // Max-heap to store {current_penalty, original_size, num_pieces}6        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);7 8        for (int num : nums) {9            pq.offer(new int[]{num, num, 1});10        }11 12        for (int i = 0; i < maxOperations; i++) {13            int[] top = pq.poll();14            int original_size = top[1];15            int num_pieces = top[2];16 17            if (original_size == num_pieces) { // Cannot split further18                pq.offer(top); // Put it back19                i--; // This operation was wasted, but constraints likely prevent this.20                continue;21            }22 23            int num_pieces_new = num_pieces + 1;24            // new_penalty = ceil(original_size / num_pieces_new)25            int new_penalty = (original_size + num_pieces_new - 1) / num_pieces_new;26            27            pq.offer(new int[]{new_penalty, original_size, num_pieces_new});28        }29 30        return pq.peek()[0];31    }32}

• Define a search range for the penalty, low = 1 and high = max(nums).
• While low <= high:
  • Pick a mid value as the candidate penalty.
  • Check if this mid penalty is achievable using a helper function isPossible(penalty).
  • The isPossible(penalty) function works as follows:
    • Initialize operations_needed = 0.
    • For each num in nums, the number of operations to make all its pieces at most penalty is (num - 1) / penalty. Add this to operations_needed.
    • Return operations_needed <= maxOperations.
  • If isPossible(mid) is true, it means mid is a valid penalty. We store it as a potential answer and try for an even smaller penalty by setting high = mid - 1.
  • Otherwise, mid is too small, and we need to allow a larger penalty, so we set low = mid + 1.
• Return the last valid penalty found.

The range of possible answers for the minimum penalty is from 1 to the maximum number of balls in any single bag, max(nums). Let's denote this range as [low, high]. We can perform a binary search on this range. For a given candidate penalty mid, we need to check if it's achievable.

To check if a penalty p is achievable, we calculate the total number of operations required to ensure no bag has more than p balls. For each bag nums[i]:

• If nums[i] <= p, no operations are needed for this bag.
• If nums[i] > p, we must split it. To make all resulting pieces have size at most p, we need to split it into k = ceil(nums[i] / p) pieces. This requires k - 1 operations.

The number of operations for a bag nums[i] can be calculated with integer division as (nums[i] - 1) / p. This formula works even when nums[i] <= p, as it correctly yields 0.

We sum these required operations for all bags. If the total sum is less than or equal to maxOperations, then the penalty p is achievable.

The binary search proceeds as follows:

1. Set low = 1, high = max(nums) (or a sufficiently large number like 10^9). ans can be initialized to high.
2. While low <= high: a. Calculate mid = low + (high - low) / 2. b. Check if penalty mid is achievable by calculating the required operations. c. If it is achievable (required_ops <= maxOperations), it means we might be able to do even better with a smaller penalty. So, we record mid as a potential answer (ans = mid) and try the lower half of the search space (high = mid - 1). d. If it's not achievable, the penalty mid is too small. We need to allow a larger penalty. So, we search in the upper half (low = mid + 1).

The final value of ans will be the minimum possible penalty.

1class Solution {2    public int minimumSize(int[] nums, int maxOperations) {3        int low = 1;4        int high = 0;5        for (int num : nums) {6            high = Math.max(high, num);7        }8 9        int minPenalty = high;10 11        while (low <= high) {12            int mid = low + (high - low) / 2;13            if (mid == 0) { // Penalty must be at least 114                low = 1;15                continue;16            }17 18            if (isPossible(nums, maxOperations, mid)) {19                minPenalty = mid;20                high = mid - 1; // Try for a smaller penalty21            } else {22                low = mid + 1; // Penalty is too small, need to increase23            }24        }25        return minPenalty;26    }27 28    // Helper function to check if a penalty 'p' is possible29    private boolean isPossible(int[] nums, int maxOperations, int p) {30        long requiredOps = 0;31        for (int num : nums) {32            // For a bag of size 'num', we need to split it into pieces of size at most 'p'.33            // Number of operations = ceil(num / p) - 1.34            // Using integer division, this is (num - 1) / p.35            requiredOps += (long)(num - 1) / p;36        }37        return requiredOps <= maxOperations;38    }39}