Minimum Limit of Balls in a Bag

lcpy gen -n 1760   # by problem number
lcpy gen -s minimum_limit_of_balls_in_a_bag   # by problem name

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.

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.

class Solution:
    # Time: O(n * log(max(nums)))
    # Space: O(1)
    def minimum_size(self, nums: list[int], max_operations: int) -> int:
        def ops_needed(penalty: int) -> int:
            return sum((n - 1) // penalty for n in nums)

        lo, hi = 1, max(nums)
        while lo < hi:
            mid = (lo + hi) // 2
            if ops_needed(mid) <= max_operations:
                hi = mid
            else:
                lo = mid + 1
        return lo