Smallest Even Multiple

Difficulty: Easy | Acceptance: 88.30% | Paid: No Topics: Math, Number Theory

Given a positive integer n, return the smallest positive multiple of n that is also even.

• Examples
• Constraints
• Brute Force
• Mathematical (LCM)
• Direct Formula
• Bit Manipulation

Examples

Example 1

Input: n = 5
Output: 10
Explanation: The smallest positive multiple of 5 that is also even is 10.

Example 2

Input: n = 6
Output: 6
Explanation: The smallest positive multiple of 6 that is also even is 6 itself.

Constraints

1 <= n <= 150

Brute Force

Intuition Start from n and keep adding n until we find an even number.

Steps

• If n is even, return n
• Otherwise, keep adding n to the result until it becomes even

class Solution:
    def smallestEvenMultiple(self, n: int) -> int:
        result = n
        while result % 2 != 0:
            result += n
        return result

Complexity

• Time: O(1) - at most one iteration needed
• Space: O(1)
• Notes: Simple but not the most efficient

Mathematical (LCM)

Intuition The smallest even multiple of n is the LCM of n and 2.

Steps

• Calculate LCM(n, 2) = (n * 2) / GCD(n, 2)
• If n is even, GCD(n, 2) = 2, so LCM = n
• If n is odd, GCD(n, 2) = 1, so LCM = 2n

import math

class Solution:
    def smallestEvenMultiple(self, n: int) -> int:
        return (n * 2) // math.gcd(n, 2)

Complexity

• Time: O(log(min(n, 2))) = O(1) for GCD calculation
• Space: O(1)
• Notes: Uses the mathematical relationship between LCM and GCD

Direct Formula

Intuition If n is even, n itself is the answer. If n is odd, we need to multiply by 2.

Steps

• Check if n is even using modulo operation
• If even, return n
• If odd, return 2 * n

class Solution:
    def smallestEvenMultiple(self, n: int) -> int:
        return n if n % 2 == 0 else 2 * n

Complexity

• Time: O(1)
• Space: O(1)
• Notes: Most efficient and readable approach

Bit Manipulation

Intuition Check the least significant bit to determine if n is even or odd.

Steps

• Use bitwise AND with 1 to check if n is odd
• If n & 1 is 0, n is even, return n
• If n & 1 is 1, n is odd, return n << 1 (n * 2)

class Solution:
    def smallestEvenMultiple(self, n: int) -> int:
        return n if (n & 1) == 0 else n << 1

Complexity

• Time: O(1)
• Space: O(1)
• Notes: Uses bitwise operations which can be slightly faster than modulo