Problem

Given a positive integer n, return _the smallest positive integer that is a multiple ofboth _2 andn.

Examples

Example 1

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    Input: n = 5
    Output: 10
    Explanation: The smallest multiple of both 5 and 2 is 10.

Example 2

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    Input: n = 6
    Output: 6
    Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

Constraints

  • 1 <= n <= 150

Solution

Method 1 – Simple Math (LCM)

Intuition

The smallest number that is a multiple of both n and 2 is their least common multiple (LCM). For any integer n, LCM(n, 2) is n if n is even, or 2n if n is odd.

Approach

  1. If n is even, return n.
  2. If n is odd, return 2 * n.

Code

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

Complexity

  • ⏰ Time complexity: O(1) — Constant time arithmetic.
  • 🧺 Space complexity: O(1) — No extra space used.