1283. Find the Smallest Divisor Given a Threshold (M)

Find the Smallest Divisor Given a Threshold (M)

题目

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

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题意

给定一个整数数组和一个阈值，要求找到一个最小的除数，使得数组中每个整数除以这个除数的商之和小于等于阈值。

思路

因为值域固定，可以用二分法夹出答案。

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代码实现

Java

class Solution {
    public int smallestDivisor(int[] nums, int threshold) {
        int left = 1, right = 1000000;
        while (left < right) {
            int mid = (right - left) / 2 + left;
            int sum = 0;
            for (int num : nums) {
                sum += Math.ceil(1.0 * num / mid);
            }
            if (sum <= threshold) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return right;
    }
}