How do you solve Smallest Even Multiple on LeetCode?

Smallest Even Multiple (LeetCode #2413) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The smallest even multiple of a number can be derived directly based on whether the number is even or odd.

What is the brute force solution for Smallest Even Multiple?

The brute force approach for Smallest Even Multiple is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each integer starting from 1 to find the smallest number that is a multiple of both 2 and n. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Smallest Even Multiple?

Smallest Even Multiple uses the following concepts: Math, Number Theory. The recommended patterns to study are: Mathematical Properties, Conditionals.

What are the interview tips for Smallest Even Multiple?

Always consider if a mathematical property can simplify the problem before diving into coding. Explain your thought process clearly; interviewers appreciate understanding how you arrive at a solution. Practice dry-running your code with examples to ensure correctness before submitting.

What are common mistakes when solving Smallest Even Multiple?

Students may forget to check if n is even or odd, leading to incorrect results. Some may attempt to implement the brute-force solution without realizing a direct mathematical approach exists.

#2413

Smallest Even Multiple

Easy

MathNumber TheoryMathematical PropertiesConditionals

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1)Space O(1)

The optimal solution leverages the fact that the smallest even multiple of n can be derived directly. If n is even, the answer is n; if n is odd, the answer is n multiplied by 2. This avoids unnecessary iterations.

⚙️

Algorithm

3 steps

1Step 1: Check if n is even.
2Step 2: If n is even, return n.
3Step 3: If n is odd, return n multiplied by 2.

solution.py4 lines

1# Full working Python code
2
3def smallest_even_multiple(n):
4    return n if n % 2 == 0 else n * 2

ℹ

Complexity note: This is constant time complexity since we only perform a couple of arithmetic operations and a modulus check, regardless of the size of n.

1The smallest even multiple of a number can be derived directly based on whether the number is even or odd.
2Understanding the properties of numbers (like even and odd) can lead to significant optimizations.

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