The `abs()`

function returns the absolute value of a number, which is the number's positive value. Essentially, it measures the distance of a number from zero on the number line.

We need to create our own function `abs()`

that mimics this functionality without using Java's built-in `abs()`

function.

abs(-1.217197940) ➞ 1.21719794 abs(-544.0) ➞ 544 abs(250.3) ➞ 250.3 abs(0) ➞ 0 abs(-0.14) ➞ 0.14

To solve this problem, we need to understand the core challenge: determining whether a number is negative and converting it to its positive counterpart if it is.

We can achieve this by using a simple conditional check:

- If the number is negative, return its positive counterpart by multiplying it by -1.
- If the number is positive or zero, return it as is.

A naive solution involves checking if the number is less than zero and then converting it to positive if it is. This approach is straightforward and efficient with a time complexity of O(1).

The naive solution is already optimal for this problem since it operates in constant time O(1). There are no further optimizations needed.

- Check if the number is less than zero.
- If it is, multiply the number by -1 to make it positive.
- If it is not, return the number as is.

```
public class AbsoluteValue {
// Function to return the absolute value of a number
public static double abs(double n) {
// Check if the number is negative
if (n < 0) {
// If negative, return the positive counterpart
return -n;
} else {
// If positive or zero, return the number as is
return n;
}
}
// Main method to test the abs function
public static void main(String[] args) {
System.out.println(abs(-1.217197940)); // ➞ 1.21719794
System.out.println(abs(-544.0)); // ➞ 544
System.out.println(abs(250.3)); // ➞ 250.3
System.out.println(abs(0)); // ➞ 0
System.out.println(abs(-0.14)); // ➞ 0.14
}
}
```

The time complexity of this solution is O(1) because it involves a single conditional check and a possible multiplication, both of which are constant-time operations.

The space complexity is also O(1) as no additional space is required beyond a few variables.

Consider the following edge cases:

- Zero:
`abs(0) ➞ 0`

- Negative numbers:
`abs(-0.14) ➞ 0.14`

- Large negative numbers:
`abs(-1000000) ➞ 1000000`

Our algorithm handles these cases effectively by design.

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

- Simple positive and negative numbers
- Zero
- Floating-point numbers
- Large magnitude numbers

Use a testing framework like JUnit for automated testing.

When approaching such problems:

- Break down the problem into smaller, manageable parts.
- Think about edge cases and how your solution handles them.
- Practice similar problems to improve your problem-solving skills.

Recreating the `abs()`

function is a simple yet fundamental exercise in understanding conditional logic and basic operations. Mastering such problems helps build a strong foundation in programming.