Leetcode_1283_Find the Smallest Divisor Given a Threshold

[lihe]

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

算法思路：

​ 记cal(x)为以x为除数，数组里每个数除以x的累加和，易知cal(x)是一个单调递减函数，可以利用函数的单调性，用二分法快速定位x大小。

class Solution{
    public int smallestDivisor(int[] nums, int threshold){
        int lo = 1, hi = 1000000;

        while (lo < hi){
            int mid = (lo + hi) >> 1;
            int result = cal(nums, mid);
            
            if(result > threshold)
                lo = mid + 1;
            else
                hi = mid;
        }
        return lo;
    }
    
    private int cal(int[] nums, int div){
        int sum = 0;
        for(int n : nums){
            sum += Math.ceil((double)n / (double)div);  // 向上取整
            //if(n % div != 0) sum += 1;
        }
        return (int)sum;
    }
}