Skip to content

Latest commit

 

History

History
150 lines (120 loc) · 3.61 KB

File metadata and controls

150 lines (120 loc) · 3.61 KB

中文文档

Description

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor, divide all the array by it, and sum the division's result. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

 

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
If the divisor is 4 we can get a sum of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2). 

Example 2:

Input: nums = [44,22,33,11,1], threshold = 5
Output: 44

Example 3:

Input: nums = [21212,10101,12121], threshold = 1000000
Output: 1

Example 4:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 1 <= nums[i] <= 106
  • nums.length <= threshold <= 106

Solutions

Python3

class Solution:
    def smallestDivisor(self, nums: List[int], threshold: int) -> int:
        left, right = 1, 1000000
        while left < right:
            mid = (left + right) >> 1
            s = 0
            for num in nums:
                s += (num + mid - 1) // mid
            if s <= threshold:
                right = mid
            else:
                left = mid + 1
        return left

Java

class Solution {
    public int smallestDivisor(int[] nums, int threshold) {
        int left = 1, right = 1000000;
        while (left < right) {
            int mid = (left + right) >> 1;
            int s = 0;
            for (int num : nums) {
                s += (num + mid - 1) / mid;
            }
            if (s <= threshold) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int smallestDivisor(vector<int>& nums, int threshold) {
        int left = 1, right = 1000000;
        while (left < right) {
            int mid = left + right >> 1;
            int s = 0;
            for (int& num : nums) {
                s += (num + mid - 1) / mid;
            }
            if (s <= threshold) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func smallestDivisor(nums []int, threshold int) int {
	left, right := 1, 1000000
	for left < right {
		mid := (left + right) >> 1
		s := 0
		for _, num := range nums {
			s += (num + mid - 1) / mid
		}
		if s <= threshold {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

...