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Weekly Contest 228 Q3
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1760. Minimum Limit of Balls in a Bag

中文文档

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. 
    <ul>
	<li>For example, a bag of <code>5</code> balls can become two new bags of <code>1</code> and <code>4</code> balls, or two new bags of <code>2</code> and <code>3</code> balls.</li>
</ul>
</li>

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 1

Python3

class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        def check(mx: int) -> bool:
            return sum((x - 1) // mx for x in nums) <= maxOperations

        return bisect_left(range(1, max(nums)), True, key=check) + 1

Java

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int left = 1, right = 0;
        for (int x : nums) {
            right = Math.max(right, x);
        }
        while (left < right) {
            int mid = (left + right) >> 1;
            long cnt = 0;
            for (int x : nums) {
                cnt += (x - 1) / mid;
            }
            if (cnt <= maxOperations) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int minimumSize(vector<int>& nums, int maxOperations) {
        int left = 1, right = *max_element(nums.begin(), nums.end());
        while (left < right) {
            int mid = (left + right) >> 1;
            long long cnt = 0;
            for (int x : nums) {
                cnt += (x - 1) / mid;
            }
            if (cnt <= maxOperations) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func minimumSize(nums []int, maxOperations int) int {
	r := slices.Max(nums)
	return 1 + sort.Search(r, func(mx int) bool {
		mx++
		cnt := 0
		for _, x := range nums {
			cnt += (x - 1) / mx
		}
		return cnt <= maxOperations
	})
}

TypeScript

function minimumSize(nums: number[], maxOperations: number): number {
    let left = 1;
    let right = Math.max(...nums);
    while (left < right) {
        const mid = (left + right) >> 1;
        let cnt = 0;
        for (const x of nums) {
            cnt += ~~((x - 1) / mid);
        }
        if (cnt <= maxOperations) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} maxOperations
 * @return {number}
 */
var minimumSize = function (nums, maxOperations) {
    let left = 1;
    let right = Math.max(...nums);
    while (left < right) {
        const mid = (left + right) >> 1;
        let cnt = 0;
        for (const x of nums) {
            cnt += ~~((x - 1) / mid);
        }
        if (cnt <= maxOperations) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};