Problem 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.
Key Insights
- We need to find a divisor that minimizes the sum of the results of the divisions.
- The result of each division is calculated using the ceiling function, which means we need to consider how to effectively compute the sum of ceilings for each divisor.
- A binary search approach can be utilized to efficiently find the smallest divisor that meets the threshold requirement.
Space and Time Complexity
Time Complexity: O(n log m), where n is the number of elements in nums and m is the maximum value in nums.
Space Complexity: O(1), as we only use a constant amount of additional space.
Solution
To solve the problem, we can use binary search to find the smallest divisor. The range of potential divisors is from 1 to the maximum number in the array. For each divisor, we calculate the sum of the ceiling of the divisions of each number in the array by the divisor. If the sum exceeds the threshold, we need to increase the divisor; if it's within the threshold, we attempt to decrease it. This process continues until we hone in on the smallest valid divisor.