LeetCode 1760. Minimum Limit of Balls in a Bag - 二分查找（Binary Search）系列题19

Posted 2022-01-13 CP Coding

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 an you should return 2.

Example 3:

Input: nums = [7,17], maxOperations = 2
Output: 7

Constraints:

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

题目大意：给定一个数组nums, 数组的长度n表示有n个袋子，第i个元素表示第i个袋子里装有nums[i]个球。现在允许你对袋子的球动手脚，动手脚的次数不能超过给定的上限maxOperations。所谓动手脚就是说每次你可以任意挑一个袋子，然后把袋子的球分装到两个袋子里（可以任意分装但要求每个袋子必须都有至少一个球）。到最后剩球最多的那个袋子的球的个数是对你的惩罚数，现在你的任务是尽可能地使惩罚数最小，返回这个最小惩罚数。

搞懂题目就会发现其实这题跟LeetCode 1011. Capacity To Ship Packages Within D Days很像。最理想的状态就是最后每个袋子里都只有一个球即惩罚数就为1，最不理想的状态就是没动任何手脚那惩罚数就原数组里的最大值，因此我们要做的就是从1到数组最大值maxNum的区间范围内找到一个最小的且满足条件的值，这很自然地想到了二分查找法。

取[1, maxNum]的中间值mid=(1 + maxNum) // 2，计算当惩罚数为mid时要动几次手脚才能使所有袋子的球个数都小于等于mid，要是动手脚次数小于等于maxOperations说明可以使惩罚数小于等于mid，则继续在左半区间查找更小的可能的惩罚数；要是动手脚次数大于maxOperations说明无法使惩罚数小于等于mid，则继续在右半区间查找更大的可能的惩罚数。

如何判断惩罚数为mid 时需要几次操作才能使所有袋子球的个数小于等于mid？遍历整个数组nums，如果球个数num小于等于mid则为0次，如果球个数num大于mid，则需要的次数为(num - 1) // mid。

class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        l, r = 1, max(nums)
        
        while l <= r:
            mid = (l + r) // 2
            total = 0
            for num in nums:
                if num > mid:
                    total += (num - 1) // mid
            if total > maxOperations:
                l = mid + 1
            else:
                r = mid - 1
        
        return l