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

Solution: Binary Search

• Time complexity: O(nlogn)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} maxOperations
 * @return {number}
 */
const minimumSize = function (nums, maxOperations) {
  let left = 1;
  let right = 10 ** 9;

  while (left < right) {
    const mid = Math.floor((left + right) / 2);
    const operation = nums.reduce((sum, num) => {
      return sum + Math.floor((num - 1) / mid);
    }, 0);

    operation > maxOperations ? (left = mid + 1) : (right = mid);
  }
  return left;
};