Problem Description

Given a positive integer n, you need to find the smallest positive integer that is a multiple of both 2 and n.

In other words, you're looking for the least common multiple (LCM) of 2 and n. This number must be divisible by both 2 and n, and it should be the smallest such positive integer.

For example:

• If n = 5, the smallest number that is a multiple of both 2 and 5 is 10
• If n = 6, the smallest number that is a multiple of both 2 and 6 is 6 itself (since 6 is already even)

The solution leverages a simple mathematical observation:

• When n is even (divisible by 2), then n itself is already a multiple of both 2 and n, so the answer is n
• When n is odd, the smallest multiple of both 2 and n is n * 2

This is because when n is odd, it has no factor of 2, so we need to multiply it by 2 to make it even while still being a multiple of n.

Intuition

To find the smallest positive integer that is a multiple of both 2 and n, we need to think about what makes a number divisible by both values.

A number that is divisible by both 2 and n must contain all the prime factors of both numbers. Since 2 is a prime number itself, our result must be even (contain at least one factor of 2).

Let's consider two cases:

1. When n is even: If n is already even, it means n already contains 2 as a factor. For example, if n = 6, then n = 2 × 3. Since n already has the factor 2, it's automatically divisible by both 2 and itself. Therefore, n is the smallest such multiple.

2. When n is odd: If n is odd, it doesn't contain 2 as a factor. For example, if n = 5, then to create a number divisible by both 2 and 5, we need to include both factors. The smallest way to do this is to multiply n by 2, giving us 2n. This ensures our number has the factor 2 (making it even) while maintaining all factors of n.

This leads us to a simple decision: check if n is even using the modulo operator (n % 2). If the remainder is 0, then n is even and already our answer. Otherwise, we need to multiply n by 2.

Learn more about Math patterns.

Solution Approach

The implementation follows directly from our mathematical understanding of the least common multiple (LCM) of 2 and n.

The algorithm is straightforward:

1. Check if n is even: We use the modulo operator n % 2 to determine if n is divisible by 2.

   • If n % 2 == 0, then n is even

2. Return the appropriate result:

   • If n is even, the LCM of 2 and n is simply n itself
   • If n is odd, the LCM of 2 and n is n * 2

The solution can be implemented in a single line using a conditional expression:

return n if n % 2 == 0 else n * 2

This approach works because:

• For even n: Since n = 2k for some integer k, both 2 and n divide n
• For odd n: Since gcd(2, n) = 1 when n is odd, the LCM formula gives us LCM(2, n) = (2 * n) / gcd(2, n) = 2n / 1 = 2n

The time complexity is O(1) as we only perform a constant number of operations (one modulo check and possibly one multiplication). The space complexity is also O(1) as we don't use any additional data structures.

Example Walkthrough

Let's walk through the solution with two examples to illustrate both cases:

Example 1: n = 5 (odd number)

1. First, we check if n is even by calculating n % 2:

   • 5 % 2 = 1 (remainder is 1, so 5 is odd)

2. Since n is odd, we need to multiply it by 2:

   • Result = 5 * 2 = 10

3. Verification:

   • Is 10 divisible by 2? Yes: 10 ÷ 2 = 5
   • Is 10 divisible by 5? Yes: 10 ÷ 5 = 2
   • Is 10 the smallest such number? Yes, because:
     • 5 is not divisible by 2
     • Any smaller multiple of 5 (which is just 5 itself) isn't even

Example 2: n = 6 (even number)

1. First, we check if n is even by calculating n % 2:

   • 6 % 2 = 0 (remainder is 0, so 6 is even)

2. Since n is even, it already contains the factor 2:

   • Result = 6

3. Verification:

   • Is 6 divisible by 2? Yes: 6 ÷ 2 = 3
   • Is 6 divisible by 6? Yes: 6 ÷ 6 = 1
   • Is 6 the smallest such number? Yes, because any positive multiple of 6 would be at least 6

The algorithm correctly identifies that odd numbers need to be doubled to become multiples of 2, while even numbers are already multiples of both 2 and themselves.

Solution Implementation

Time and Space Complexity

Time Complexity: O(1)

The function performs a constant number of operations regardless of the input size:

• One modulo operation (n % 2)
• One comparison (== 0)
• One conditional return with either direct return of n or one multiplication (n * 2)

All these operations take constant time, resulting in O(1) time complexity.

Space Complexity: O(1)

The function uses only a constant amount of extra space:

• No additional data structures are created
• Only the input parameter n and the return value are stored
• The space used does not grow with the input size

Therefore, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Overthinking the Mathematical Formula

Many developers might try to implement the full LCM formula using GCD (Greatest Common Divisor):

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def smallestEvenMultiple(n):
    return (2 * n) // gcd(2, n)  # Unnecessarily complex

Why it's a pitfall: While mathematically correct, this approach is overly complex for this specific problem where one number is always 2. The GCD calculation adds unnecessary overhead.

Solution: Recognize that when dealing with 2 specifically:

• GCD(2, even_number) = 2
• GCD(2, odd_number) = 1

This simplifies to the straightforward even/odd check.

2. Using Bitwise Operations Without Clear Understanding

Some might attempt to optimize using bitwise operations:

def smallestEvenMultiple(n):
    return n << (n & 1)  # Clever but unclear

Why it's a pitfall: While this works (it left-shifts n by 1 if n is odd, 0 if even), it sacrifices readability for marginal performance gain. The bitwise approach can confuse team members and make debugging harder.

Solution: Stick with the clear, readable approach using modulo unless performance profiling specifically shows this function as a bottleneck.

3. Not Handling Edge Cases in Interview Settings

In an interview, candidates might forget to clarify constraints:

def smallestEvenMultiple(n):
    # What if n is 0 or negative?
    return n if n % 2 == 0 else n * 2

Why it's a pitfall: The problem states n is positive, but in real-world scenarios or extended versions of the problem, you might need to handle:

• n = 0 (mathematically undefined LCM)
• Negative numbers
• Very large numbers causing overflow

Solution: Add input validation if the problem constraints aren't guaranteed:

def smallestEvenMultiple(n):
    if n <= 0:
        raise ValueError("n must be a positive integer")
    return n if n % 2 == 0 else n * 2

4. Misunderstanding the Problem Statement

Some might interpret "smallest multiple" as finding factors instead:

def smallestEvenMultiple(n):
    # Wrong: trying to find smallest factor
    for i in range(2, n + 1):
        if i % 2 == 0 and n % i == 0:
            return i
    return n * 2

Why it's a pitfall: This finds the smallest even divisor of n, not the LCM of 2 and n.

Solution: Carefully read the problem - we need a number divisible by BOTH 2 and n, not a number that divides n.