1760. Minimum Limit of Balls in a Bag

leetcode_1760.java

/**
1760. Minimum Limit of Balls in a Bag

Medium

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

*/

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int n = nums.length;
        int bad = 0;
        int good = (int)Math.pow(10, 9);
        while (good - bad > 1) {
            int m = bad + (good - bad) / 2;
            int curr_ops = 0;
            for (int i = 0; i < n; i++) {
                if (nums[i] > m) {
                    curr_ops += (nums[i] - 1) / m;
                }
                if (curr_ops > maxOperations) {
                    break;
                }
            }
            if (curr_ops > maxOperations) {
                bad = m;
            } else {
                good = m;
            }
        }
        return good;
    }
}