Formatted question description: https://leetcode.ca/all/2413.html

2413. Smallest Even Multiple

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: Greatest Common Divisor of Strings, Three Divisors, Find Greatest Common Divisor of Array.

Problem

Given a positive integer n, return the smallest positive integer that is a multiple of **both 2 and n.   **Example 1:

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


Example 2:

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 (Java, C++, Python)

• class Solution {
public int smallestEvenMultiple(int n) {
int k=2*n;
for(int i=2*n;i>=n;i--)
{
if(i%n==0&&i%2==0)
k=i;
}
return k;
}
}

############

class Solution {
public int smallestEvenMultiple(int n) {
return n % 2 == 0 ? n : n * 2;
}
}

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

############

# 2413. Smallest Even Multiple
# https://leetcode.com/problems/smallest-even-multiple/

class Solution:
def smallestEvenMultiple(self, n: int) -> int:
for x in range(1, n * 2 + 1):
if x % 2 == 0 and x % n == 0:
return x

return -1


• class Solution {
public:
int smallestEvenMultiple(int n) {
return n % 2 == 0 ? n : n * 2;
}
};

• func smallestEvenMultiple(n int) int {
if n%2 == 0 {
return n
}
return n * 2
}

• function smallestEvenMultiple(n: number): number {
return n % 2 === 0 ? n : n * 2;
}


• impl Solution {
pub fn smallest_even_multiple(n: i32) -> i32 {
if n % 2 == 0 {
return n;
}
n * 2
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).