1886. Minimum Limit Of Balls In A Bag¶

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1886. 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 <= 105
• 1 <= maxOperations, nums[i] <= 109

Solution¶

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int n = nums.length;
        int low = 1, high = (int)(1e9);
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (check(mid, nums, maxOperations)) {
                ans = mid;
                high = mid - 1;
            }
            else low = mid + 1;
        }
        return ans;
    }

    private boolean check(int mid, int arr[], int k) {
        int n = arr.length;
        long count = 0;
        for (int i = 0; i < n; i++) {
            if (count > k) return false;
            count += (arr[i] - 1) / mid;
        }
        return count <= k;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here