How do you solve Closest Divisors on LeetCode?

Closest Divisors (LeetCode #1362) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(sqrt(n)) time complexity and O(1) space complexity. Key insight: Divisors can be found efficiently by only checking up to the square root of the number.

What is the brute force solution for Closest Divisors?

The brute force approach for Closest Divisors is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of integers to see if their product equals num + 1 or num + 2. This is straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(sqrt(n)).

What algorithm or data structure is used to solve Closest Divisors?

Closest Divisors uses the following concepts: Math. The recommended patterns to study are: Math, Two Pointers.

What are the interview tips for Closest Divisors?

Always consider edge cases, such as very small or very large input values. Explain your thought process clearly as you write code; it helps interviewers understand your reasoning. Practice dry runs of your code with different inputs to ensure correctness before finalizing your solution.

What are common mistakes when solving Closest Divisors?

Not optimizing the search for divisors, leading to unnecessary computations. Failing to handle the case where the closest divisors are found for both num + 1 and num + 2.

#1362

Closest Divisors

Medium

MathMathTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(sqrt(n))
Space	O(1)	O(1)

💡

Intuition

Time O(sqrt(n))Space O(1)

The optimal approach leverages the fact that we only need to check divisors up to the square root of the numbers. This significantly reduces the number of checks needed to find the closest divisors.

⚙️

Algorithm

3 steps

1Step 1: Calculate num + 1 and num + 2.
2Step 2: For each of these two values, iterate from 1 to the square root of the number to find divisors.
3Step 3: For each divisor found, calculate its pair and track the pair with the smallest absolute difference.

solution.py13 lines

1import math
2
3def closestDivisors(num):
4    def findClosestDivisors(n):
5        closest = (1, n)
6        for i in range(1, int(math.sqrt(n)) + 1):
7            if n % i == 0:
8                j = n // i
9                if abs(i - j) < abs(closest[0] - closest[1]):
10                    closest = (i, j)
11        return closest
12
13    return findClosestDivisors(num + 1) if findClosestDivisors(num + 1)[0] <= findClosestDivisors(num + 2)[0] else findClosestDivisors(num + 2)

ℹ

Complexity note: This complexity is much better because we only check divisors up to the square root of the numbers, which drastically reduces the number of iterations needed.

1Divisors can be found efficiently by only checking up to the square root of the number.
2The closest divisors will have the smallest absolute difference, which can be tracked during the search.

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