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

Code

基本概念同：Binary Search 101

Time Complexity: $O(n\cdot \log 1e9)$, Space Complexity: $O(1)$

class Solution {
public:
    int minimumSize(vector<int>& nums, int maxOperations) {
        int l = 1, r = 1e9;
 
        while(l <= r) {
            int m = l + (r - l) / 2;
            if(canDivide(m, nums, maxOperations)) {
                r = m - 1;
            } else {
                l = m + 1;
            }
        }
 
        return l;
    }
 
    bool canDivide(int maxPenalty, vector<int>& nums, int maxOperations) {
        int penalty;
        int operation = 0;
        for(auto&n : nums) {
            if(n > maxPenalty) {
                operation += (n / maxPenalty) - 1;
                if ((n % maxPenalty) > 0) operation++;
            }
            if(operation > maxOperations) return false;
        }
 
        return true;
    }
};

注意在計算要切分多少次時，可以直接計算 operation += ((n - 1) / maxPenalty);。

Source

• Minimum Limit of Balls in a Bag - LeetCode