1760. Minimum Limit of Balls in a Bag

Problem

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 an you should return 2.

Example 3:

Input: nums = [7,17], maxOperations = 2
Output: 7

  Constraints:

• 1 <= nums.length <= 10^5

• 1 <= maxOperations, nums[i] <= 10^9

Solution (Java)

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int left = 1;
        int right = 1_000_000_000;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (operations(nums, mid) > maxOperations) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }

    private int operations(int[] nums, int mid) {
        int operations = 0;
        for (int num : nums) {
            operations += (num - 1) / mid;
        }
        return operations;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).