Find the smallest divisor given a threshold

Problem

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).

Some questions to ask

• What to return if there isn’t a valid divisor?
• Is empty array valid input?

Approach

Smallest ‘useful’ divisor is 1, largest ‘useful’ divisor is the maximum value in the array.

Find the smallest divisor that satisfies the constraint using binary search, starting with the range [1, x] where x = max(nums).

def smallestDivisor(nums: List[int], threshold: int) -> int:
    """Return smallest divisor such that sum of `nums` normalised
    with respect to the divisor is at most `threshold`."""
    
    lo = 1
    hi = max(nums)

    while lo < hi:
        mid = floor((lo + hi) / 2)
        normalisedSum = sum(ceil(num / mid) for num in nums)

        if normalisedSum <= threshold:
            # Can try smaller threshold, but keep `mid` as candidate
            hi = mid
        else:
            # Sum too large, need bigger divisor
            lo = mid + 1

    return lo

Complexity

Let x be the maximum value in nums.

• O(log(x)) time complexity, from binary search
• O(1) space complexity

GitHub: smallest_divisor_given_threshold