You are given an integer array nums where the i<sup>th</sup> bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

• Take any bag of balls and divide it into two new bags with a positive number of balls.
  • For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

Input: nums = [9], maxOperations = 2
Output: 3
Explanation: 
- Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3].
- Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3].
The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2.

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= maxOperations, nums[i] <= 10^9

Naive Approach

A brute-force solution would involve trying out every possible combination of divisions for each bag, but this would be extremely inefficient, especially given the constraints on the size of nums and the values within it.

Optimal Approach: Binary Search

The problem asks us to minimize the maximum number of balls in a bag after performing a limited number of operations. This structure strongly suggests using binary search. We can binary search for the minimum possible penalty. For a given penalty value, we check if it's possible to achieve that penalty within the allowed number of operations.

1. Define Search Space: The minimum possible penalty is 1, and the maximum possible penalty is the largest value in nums.
2. Binary Search: Pick a mid value within the search space (potential penalty).
3. Check Feasibility: For each bag, determine how many operations are needed to make the largest number of balls in the bag less than or equal to mid. This is done by calculating ceil(nums[i] / mid) - 1. Sum up these operations for all bags.
4. Adjust Search Space: If the total operations needed are less than or equal to maxOperations, then we can potentially reduce the penalty, so we move the high pointer to mid - 1. Otherwise, we need to increase the penalty, so we move the low pointer to mid + 1.
5. Return: When low > high, low will be the minimum possible penalty.

def minimumSize(nums, maxOperations):
    low = 1
    high = max(nums)

    while low <= high:
        mid = (low + high) // 2
        operationsNeeded = 0
        for num in nums:
            operationsNeeded += (num - 1) // mid  # Equivalent to ceil(num / mid) - 1

        if operationsNeeded <= maxOperations:
            high = mid - 1
        else:
            low = mid + 1

    return low

# Example Usage
nums = [2, 4, 8, 2]
maxOperations = 4
result = minimumSize(nums, maxOperations)
print(f"Minimum possible penalty: {result}")  # Output: 2

nums = [9]
maxOperations = 2
result = minimumSize(nums, maxOperations)
print(f"Minimum possible penalty: {result}")

Big O Analysis

• Time Complexity: O(N log M), where N is the number of bags (nums.length) and M is the maximum number of balls in a bag (max(nums)). The binary search takes O(log M) iterations, and in each iteration, we iterate through all the bags in O(N) time to calculate the number of operations needed.
• Space Complexity: O(1). The algorithm uses only a constant amount of extra space for variables like low, high, mid, and operationsNeeded.

Edge Cases

1. maxOperations = 0: If no operations are allowed, the minimum penalty is simply the maximum value in nums.
2. nums contains only one bag: The binary search will still work correctly.
3. nums contains very large numbers: The (num - 1) // mid calculation can handle large numbers without overflow because the intermediate values are integers.
4. All numbers in nums are small: The binary search will quickly converge to the optimal penalty.
5. maxOperations is very large: The binary search will still efficiently find the minimum penalty.