How do you solve Find the Smallest Divisor Given a Threshold on LeetCode?

Find the Smallest Divisor Given a Threshold (LeetCode #1283) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(1) space complexity. Key insight: Binary search significantly reduces the number of checks needed to find the smallest divisor.

What is the time complexity of Find the Smallest Divisor Given a Threshold?

The optimal solution for Find the Smallest Divisor Given a Threshold runs in O(n log m) time and uses O(1) space. This complexity arises because we perform a binary search over the range of divisors (log m) and for each divisor, we compute the total sum (O(n)).

What is the brute force solution for Find the Smallest Divisor Given a Threshold?

The brute force approach for Find the Smallest Divisor Given a Threshold is Brute Force, with O(n * m) time complexity and O(1) space complexity. The brute-force approach involves trying every possible divisor starting from 1 up to the maximum number in the array. For each divisor, we calculate the sum of the divisions and check if it meets the threshold. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Find the Smallest Divisor Given a Threshold?

Find the Smallest Divisor Given a Threshold uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Find the Smallest Divisor Given a Threshold?

Always consider edge cases, such as when all numbers are the same or when the threshold is very small. Explain your thought process clearly, especially when transitioning from brute-force to a more optimal solution. Practice implementing binary search on different problems to become comfortable with its application.

What are common mistakes when solving Find the Smallest Divisor Given a Threshold?

Not using ceiling division correctly, which can lead to incorrect sums. Failing to recognize that binary search can be applied to this problem, leading to inefficient brute-force solutions.

#1283

Find the Smallest Divisor Given a Threshold

Medium

ArrayBinary SearchBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

Using binary search allows us to efficiently find the smallest divisor by narrowing down the range of possible divisors based on the sum of divisions. This is much faster than checking every divisor.

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Algorithm

5 steps

1Step 1: Set the left boundary of the search to 1 and the right boundary to the maximum number in the array.
2Step 2: While left is less than or equal to right, calculate the middle divisor.
3Step 3: Calculate the total sum of divisions for the middle divisor.
4Step 4: If the sum is less than or equal to the threshold, move the right boundary to mid - 1 (to find a smaller divisor). Otherwise, move the left boundary to mid + 1.
5Step 5: The left boundary will converge to the smallest valid divisor.

solution.py11 lines

1def smallestDivisor(nums, threshold):
2    def total_division(divisor):
3        return sum(-(-num // divisor) for num in nums)
4    left, right = 1, max(nums)
5    while left < right:
6        mid = (left + right) // 2
7        if total_division(mid) > threshold:
8            left = mid + 1
9        else:
10            right = mid
11    return left

ℹ

Complexity note: This complexity arises because we perform a binary search over the range of divisors (log m) and for each divisor, we compute the total sum (O(n)).

1Binary search significantly reduces the number of checks needed to find the smallest divisor.
2Understanding how to calculate the sum of divisions efficiently is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.