This approach uses binary search to efficiently find the smallest divisor. The possible divisors can range from 1 to the maximum number in the array. For each candidate divisor, divide each number in the array, take the ceiling of the division result, sum it up, and compare against the threshold.
Binary search helps minimize the number of trials by adjusting the divisor range based on whether the threshold condition is met or not for a given divisor.
Time Complexity: O(n * log(max(nums))) where n is the number of elements in nums. Space Complexity: O(1) as we use constant additional space.
1import math
2
3def smallestDivisor(nums, threshold):
4 left, right = 1, 1000000
5 while left < right:
6 mid = (left + right) // 2
7 sum_div = sum((num + mid - 1) // mid for num in nums)
8 if sum_div > threshold:
9 left = mid + 1
10 else:
11 right = mid
12 return leftThe Python solution uses list comprehension with the binary search strategy. By calculating the ceiling in a repetitive fashion using integer math, it efficiently narrows down the potential divisor range.
This method involves a simple linear search from divisor = 1 upwards, incrementally checking each candidate as a divisor. This approach is less efficient compared to binary search but helps understand the problem in a straightforward manner. We calculate the total sum of the divisions for each divisor and stop when the sum is less or equal to the threshold. Ensuring progressively checking divisors until a valid one is found assures correctness.
Time Complexity: Potentially O(n * max(nums)). Space Complexity: O(1).
1def smallestDivisor(nums, threshold):
2 divisor = 1
3 while True:
4 if sum((num + divisor - 1) // divisor for num in nums) <= threshold:
5 return divisor
6 divisor += 1The Python approach is a direct translation into Python's loop and list-comp behaviour, trying each divisor until the threshold condition is met, returning the smallest valid divisor found.