leetcode 1283. Find the Smallest Divisor Given a Threshold

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

Each result of 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 to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

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

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Constraints:

1 <= nums.length <= 5 * 10^4
1 <= nums[i] <= 10^6
nums.length <= threshold <= 10^6

题目难度：简单题

题目思路：就是暴力穷举，暴力穷举过程中，因子逐渐自增。实际上，我们可以通过分析得到因子的上确界（大于这个值的所有因子，都将使得nums中的每一个值经过除法之后都为1），这样我们可以通过二分法，更快找到目标因子。

C++代码：

 1 class Solution {
 2 private:
 3     int sum(vector<int>& nums, int divisor) {
 4         int sum = 0;
 5         for (auto n : nums) {
 6             sum += n / divisor + (n % divisor == 0 ? 0 : 1);
 7         }
 8         return sum;
 9     }
10 public:
11     int smallestDivisor(vector<int>& nums, int threshold) {
12         int left = 1;
13         int right = nums[nums.size() - 1]; //max nums
14         while (left <= right) {
15             int mid = left + ((right - left) >> 1);
16             int n = sum(nums, mid);
17             if (n > threshold) {
18                 left = mid + 1;
19             } else {
20                 right = mid - 1;
21             }
22         }
23         return left;
24     }
25 };

注：时间复杂度$O(nlogn)$，这里假设了原数组有序，最后一个数为最大值，否则需要先找到最大值($O(n)$的时间复杂度)