<h2 id="heading-problem">Problem</h2>
<a target="_blank" href="https://leetcode.com/problems/smallest-even-multiple/">Smallest Even Multiple - LeetCode</a>
<h3 id="heading-solution-approach">Solution Approach</h3>
<ul>
<li>The term "multiple of 2" indicates that the number must be even.
</li>
<li>The term "multiple of n" indicates that the number must be a multiple of n.
</li>
<li>This problem is about finding the smallest even multiple among the multiples of n.
</li>
<li>Initially, I attempted an algorithm where I iterated by incrementing n and returned it if it was even. However, I soon realized that the second multiple of n is always an even number.
</li>
<li>Therefore, the most efficient algorithm is as follows:
<ul>
<li>Check if n is even.
</li>
<li>If n is even, return n.
</li>
<li>If n is not even, then the next multiple will always be even.
</li>
<li>Therefore, return n * 2.
</li>
</ul>
</li>
</ul>
<h2 id="heading-github-link">Github Link</h2>
<a target="_blank" href="https://github.com/eunhanlee/LeetCode_2413_SmallestEvenMultiple_Solution.git">https://github.com/eunhanlee/LeetCode_2413_SmallestEvenMultiple_Solution.git</a>
<h2 id="heading-time-complexity-o1-space-complexity-o1">Time Complexity: O(1), Space Complexity: O(1)</h2>
<pre><code class="lang-java">public class Solution {
 /**
 * Finds the smallest even multiple of the given number.
 *
 * @param n The number for which the smallest even multiple needs to be found.
 * @return The smallest even multiple of the given number.
 */
 public int smallestEvenMultiple(int n) {
 if (n % 2 == 0) {
 return n;
 } else {
 return n * 2;
 }
 }
}
</code></pre>

## Problem

[Smallest Even Multiple - LeetCode](https://leetcode.com/problems/smallest-even-multiple/)

### Solution Approach

* The term "multiple of 2" indicates that the number must be even.
    
* The term "multiple of n" indicates that the number must be a multiple of n.
    
* This problem is about finding the smallest even multiple among the multiples of n.
    
* Initially, I attempted an algorithm where I iterated by incrementing n and returned it if it was even. However, I soon realized that the second multiple of n is always an even number.
    
* Therefore, the most efficient algorithm is as follows:
    
    * Check if n is even.
        
    * If n is even, return n.
        
    * If n is not even, then the next multiple will always be even.
        
    * Therefore, return n \* 2.
        

## Github Link

[https://github.com/eunhanlee/LeetCode\_2413\_SmallestEvenMultiple\_Solution.git](https://github.com/eunhanlee/LeetCode_2413_SmallestEvenMultiple_Solution.git)

## Time Complexity: O(1), Space Complexity: O(1)

```java
public class Solution {
    /**
     * Finds the smallest even multiple of the given number.
     *
     * @param n The number for which the smallest even multiple needs to be found.
     * @return The smallest even multiple of the given number.
     */
    public int smallestEvenMultiple(int n) {
        if (n % 2 == 0) {
            return n;
        } else {
            return n * 2;
        }
    }
}
```

LeetCode 2413. Smallest Even Multiple Java Solution

Problem

Solution Approach

Github Link

Time Complexity: O(1), Space Complexity: O(1)