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1760. Minimum Limit of Balls in a Bag
Description
You are given an integer array nums
where the i^{th}
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 of1
and4
balls, or two new bags of2
and3
balls.
 For example, a bag of
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}
Solutions
Binary search.

class Solution { public int minimumSize(int[] nums, int maxOperations) { int left = 1, right = 0; for (int x : nums) { right = Math.max(right, x); } while (left < right) { int mid = (left + right) >> 1; long cnt = 0; for (int x : nums) { cnt += (x  1) / mid; } if (cnt <= maxOperations) { right = mid; } else { left = mid + 1; } } return left; } }

class Solution { public: int minimumSize(vector<int>& nums, int maxOperations) { int left = 1, right = *max_element(nums.begin(), nums.end()); while (left < right) { int mid = (left + right) >> 1; long long cnt = 0; for (int x : nums) { cnt += (x  1) / mid; } if (cnt <= maxOperations) { right = mid; } else { left = mid + 1; } } return left; } };

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

func minimumSize(nums []int, maxOperations int) int { r := slices.Max(nums) return 1 + sort.Search(r, func(mx int) bool { mx++ cnt := 0 for _, x := range nums { cnt += (x  1) / mx } return cnt <= maxOperations }) }

function minimumSize(nums: number[], maxOperations: number): number { let left = 1; let right = Math.max(...nums); while (left < right) { const mid = (left + right) >> 1; let cnt = 0; for (const x of nums) { cnt += ~~((x  1) / mid); } if (cnt <= maxOperations) { right = mid; } else { left = mid + 1; } } return left; }

/** * @param {number[]} nums * @param {number} maxOperations * @return {number} */ var minimumSize = function (nums, maxOperations) { let left = 1; let right = Math.max(...nums); while (left < right) { const mid = (left + right) >> 1; let cnt = 0; for (const x of nums) { cnt += ~~((x  1) / mid); } if (cnt <= maxOperations) { right = mid; } else { left = mid + 1; } } return left; };