# 1760. Minimum Limit of Balls in a Bag

## Description

You are given an integer array nums where the ith 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 <= 105
• 1 <= maxOperations, nums[i] <= 109

## 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;
};