Easy Java Solution!!  Constant time complexity!!
IntuitionThe question says it all. The least common multiple of a number is itself if it is divisible by 2 or is if even.Else, if  the number is odd, then the l
# Intuition
The question says it all. The least common multiple of a number is itself if it is divisible by 2 or is if even.

Else, if  the number is odd, then the lowest multiple of 2 and that number would be the product.
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
<!-- Describe your approach to solving the problem. -->
1. Checking if the number is even, then returning that one.
2. If odd,  return the product of two. 
# Complexity
- Time complexity: No loops, so $$O(1)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: No extra space used, so $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
![image.png](https://assets.leetcode.com/users/images/3051839c-cc04-4618-b8c0-b642069ee5c2_1738124509.7888484.png)


**Just a Single request:**

![image.png](https://assets.leetcode.com/users/images/12e1c7a9-a667-4995-8c8c-daaf45c40036_1738124745.0093725.png)


# Code
```java []
class Solution {
    public int smallestEvenMultiple(int n) {
        if(n%2==0) return n;
        else return n*2;
    }
}
```
['Java']