Formatted question description: https://leetcode.ca/all/1760.html

1760. Minimum Limit of Balls in a Bag

Level

Medium

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 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

Sort the array nums. Use binary search, where initially let low = 1 and high = nums[nums.length - 1]. Each time let mid be the mean of low and high, and calculate whether penalty = mid is possible. Find the minimum possible penalty during binary search, and finally return the minimum possible penalty.

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        Arrays.sort(nums);
        int length = nums.length;
        int low = 1, high = nums[length - 1];
        while (low < high) {
            int mid = (high - low) / 2 + low;
            if (isPossible(nums, maxOperations, mid))
                high = mid;
            else
                low = mid + 1;
        }
        return low;
    }

    public boolean isPossible(int[] nums, int maxOperations, int penalty) {
        int count = 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            int num = nums[i];
            if (num <= penalty)
                break;
            int operations = nums[i] / penalty - 1;
            if (nums[i] % penalty != 0)
                operations++;
            count += operations;
        }
        return count <= maxOperations;
    }
}

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