Inside the code editor we've tried to write a function that takes an array `nums`

as argument and prints to the console the **even** numbers from that array.

So when we called `printEvenNumbers(new int[]{2, 1, 0, 4, 3})`

, we expected our code to print:

```
2
0
4
```

but it seems like we made some mistakes because when we run our code, it prints:

```
even
even
```

**Assignment:**

Your task is to fix our function such that it prints the even numbers from the array.

The core challenge here is to correctly identify and print the even numbers from the given array. Even numbers are integers that are divisible by 2 without any remainder. This problem is common in scenarios where filtering specific elements from a collection is required.

Potential pitfalls include misunderstanding the condition for even numbers or incorrectly iterating through the array.

To solve this problem, we need to iterate through the array and check each element to see if it is even. If it is, we print it. A naive solution might involve unnecessary complexity or incorrect conditions, but the optimal solution is straightforward:

- Iterate through each element of the array.
- Check if the element is even using the modulus operator (%).
- If the element is even, print it.

Here is a step-by-step breakdown of the algorithm:

- Start a loop from the first element to the last element of the array.
- For each element, check if it is divisible by 2 (i.e.,
`element % 2 == 0`

). - If the condition is true, print the element.

```
public class EvenNumbers {
public static void printEvenNumbers(int[] nums) {
// Iterate through each element in the array
for (int num : nums) {
// Check if the number is even
if (num % 2 == 0) {
// Print the even number
System.out.println(num);
}
}
}
public static void main(String[] args) {
// Test the function with a sample array
printEvenNumbers(new int[]{2, 1, 0, 4, 3});
}
}
```

In this code:

- We use a for-each loop to iterate through the array.
- The condition
`num % 2 == 0`

checks if the number is even. - If the condition is true, the number is printed.

The time complexity of this solution is O(n), where n is the number of elements in the array. This is because we need to check each element once. The space complexity is O(1) as we are not using any additional space that scales with the input size.

Consider the following edge cases:

- An empty array: The function should handle this gracefully without any errors.
- An array with no even numbers: The function should not print anything.
- An array with all even numbers: The function should print all elements.

Examples:

```
printEvenNumbers(new int[]{}) // No output
printEvenNumbers(new int[]{1, 3, 5}) // No output
printEvenNumbers(new int[]{2, 4, 6}) // Output: 2 4 6
```

To test the solution comprehensively, consider using a variety of test cases:

- Simple cases with a mix of even and odd numbers.
- Edge cases as discussed above.
- Large arrays to ensure performance is acceptable.

JUnit or other testing frameworks can be used to automate these tests.

When approaching such problems:

- Clearly understand the problem requirements and constraints.
- Break down the problem into smaller, manageable parts.
- Consider edge cases and how your solution handles them.
- Write clean, readable code with comments to explain your logic.

Practice by solving similar problems and studying different algorithms to improve your problem-solving skills.

In this post, we discussed how to fix a function to print even numbers from an array. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong programming skills.

Keep practicing and exploring further to enhance your knowledge and abilities.