Given a non-negative integer **n**, check its parity and return "even" or "odd".

A number is "even" if it is divisible by 2 and "odd" otherwise.

**Example:**

Input:n= 12Output:"even"Explanation:12 is divisible by 2, so it is even

The core challenge of this problem is to determine whether a given number is even or odd. This is a fundamental problem in computer science and mathematics with applications in various fields such as algorithm design, data structures, and more.

Common applications include determining the parity of elements in an array, optimizing algorithms based on even or odd indices, and more.

Potential pitfalls include misunderstanding the modulo operation or incorrectly handling edge cases such as zero.

To solve this problem, we can use the modulo operator. The modulo operator (%) returns the remainder of a division operation. If a number **n** is divisible by 2, the remainder will be 0, indicating that the number is even. Otherwise, the number is odd.

A naive solution would involve checking the number using an if-else statement with the modulo operator:

if (n % 2 == 0) { return "even"; } else { return "odd"; }

This solution is straightforward and works well for this problem. However, it can be optimized further for readability and efficiency.

The optimized solution involves the same logic but can be written more concisely:

return (n % 2 == 0) ? "even" : "odd";

This approach is more concise and leverages the ternary operator for a cleaner implementation.

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

- Take the input number
**n**. - Check if
**n**modulo 2 is equal to 0. - If true, return "even".
- Otherwise, return "odd".

```
public class ParityChecker {
// Method to check the parity of a number
public static String checkParity(int n) {
// Using modulo operator to determine even or odd
return (n % 2 == 0) ? "even" : "odd";
}
// Main method to test the checkParity method
public static void main(String[] args) {
int n = 12;
System.out.println("The number " + n + " is " + checkParity(n));
}
}
```

The time complexity of this solution is O(1) because the modulo operation and the comparison are constant-time operations.

The space complexity is also O(1) as we are not using any additional data structures or memory.

Potential edge cases include:

**n = 0**: The output should be "even" since 0 is divisible by 2.**n = 1**: The output should be "odd" since 1 is not divisible by 2.**n = Integer.MAX_VALUE**: The output should be "odd" since the maximum integer value is not divisible by 2.

These edge cases can be tested to ensure the robustness of the solution.

To test the solution comprehensively, consider the following test cases:

**n = 0**: Expected output is "even".**n = 1**: Expected output is "odd".**n = 2**: Expected output is "even".**n = 13**: Expected output is "odd".**n = 100**: Expected output is "even".

These test cases cover a range of inputs from small to large values.

When approaching such problems, consider the following tips:

- Understand the problem statement and constraints thoroughly.
- Break down the problem into smaller, manageable parts.
- Consider edge cases and test your solution against them.
- Practice similar problems to improve your problem-solving skills.

In this blog post, we discussed how to determine the parity of a number using Java. 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 problem-solving skills in computer science.

We encourage you to practice and explore further to enhance your understanding and proficiency.

For further reading and practice, consider the following resources: