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. (For example: 7/3 = 3 and 10/2 = 5).

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

Example 2:

Input: nums = [44,22,33,11,1], threshold = 5
Output: 44

Constraints:

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

Finding the Smallest Divisor

This problem requires us to find the smallest divisor such that the sum of the divisions (ceiling value) of the input array elements is less than or equal to the given threshold. We can solve this using binary search.

Brute Force Solution

A naive solution would be to iterate through all possible divisors from 1 to the maximum element in the nums array and check if the sum of the divisions is less than or equal to the threshold. This solution has a time complexity of O(N*M), where N is the length of the nums array and M is the maximum value in the nums array.

Optimal Solution

We can use binary search to find the smallest divisor efficiently. The search space is from 1 to the maximum value in the nums array. For each divisor, we calculate the sum of the ceiling of the divisions and check if it is less than or equal to the threshold. If it is, we try a smaller divisor; otherwise, we try a larger divisor.

def smallestDivisor(nums, threshold):
    left, right = 1, max(nums)
    ans = right

    while left <= right:
        mid = (left + right) // 2
        total = sum((num + mid - 1) // mid for num in nums)

        if total <= threshold:
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans

Example

nums = [1, 2, 5, 9]
threshold = 6
print(smallestDivisor(nums, threshold))  # Output: 5

Big(O) Run-time Analysis

The binary search takes O(log M) time, where M is the maximum element in the nums array. For each divisor, we iterate through the nums array once to calculate the sum of the divisions, which takes O(N) time. Therefore, the overall time complexity is O(N log M).

Big(O) Space Usage Analysis

The space complexity is O(1) because we are only using a few variables to store the left, right, mid, and answer values. We are not using any additional data structures that scale with the input size.

Edge Cases

1. Empty Input Array: If the input array nums is empty, the problem statement doesn't explicitly define the behavior. We can assume that the problem statement asserts that the input array will not be empty, based on the problem constraints.
2. Threshold is less than array size: The problem states that nums.length <= threshold. This guarantees that a valid divisor exists. If this were not true, it might be impossible to find a divisor that satisfies the condition.
3. Large Input Values: If the input values in nums or the threshold are very large, integer overflow might occur. Using Python avoids such issues automatically due to dynamic typing and arbitrary-precision integers. In other languages like Java or C++, using long is necessary.
4. All Elements are 1: If all the elements are 1 and the threshold is large, the divisor can be 1.
5. Threshold equals array length: The smallest divisor equals the largest number in nums.

Additional Considerations

The binary search approach provides an efficient way to find the smallest divisor. The key idea is to leverage the monotonically decreasing property of the sum of divisions as the divisor increases. This property allows us to effectively narrow down the search space and find the optimal divisor in logarithmic time concerning the range of possible divisors.