The absolute difference is a measure of the difference between two numbers but expressed as a positive number.
In other words, it’s taking the difference between two values (x - y
) and then calculating the absolute value of the result.
Create a function that takes in two numbers a
and b
and returns their absolute difference.
Examples:
absDiff(3, 0) ➞ 3 absDiff(0, 3) ➞ 3 absDiff(10, -1) ➞ 11 absDiff(-5, -7) ➞ 2 absDiff(6, -6) ➞ 12 absDiff(1.5, 1.5) ➞ 0
Note:
Do not use the abs()
function provided by Java (it will defeat the purpose of the challenge)
The core challenge of this problem is to compute the absolute difference between two numbers without using the built-in abs()
function. The absolute difference is always a positive number, regardless of the order of the inputs.
This problem is significant in various applications such as calculating distances, error margins, and differences in data values. A common pitfall is to forget that the result must always be positive.
To solve this problem, we need to consider the following:
a
is less than b
, the absolute difference is b - a
.a
is greater than or equal to b
, the absolute difference is a - b
.We can implement this logic using an if
statement to check the condition and return the appropriate result.
A naive solution would involve checking the condition and returning the difference accordingly. This approach is straightforward and has a time complexity of O(1), which is optimal for this problem.
The naive solution is already optimal for this problem. However, we can further simplify the code by using the ternary operator to make it more concise.
Here is a step-by-step breakdown of the algorithm:
a
is less than b
.b - a
.a - b
.Below is the Java implementation of the solution:
public class AbsoluteDifference {
// Function to calculate the absolute difference between two numbers
public static double absDiff(double a, double b) {
// Check if a is less than b
if (a < b) {
return b - a; // Return b - a if a is less than b
} else {
return a - b; // Return a - b if a is greater than or equal to b
}
}
// Main method to test the function
public static void main(String[] args) {
// Test cases
System.out.println(absDiff(3, 0)); // ➞ 3
System.out.println(absDiff(0, 3)); // ➞ 3
System.out.println(absDiff(10, -1)); // ➞ 11
System.out.println(absDiff(-5, -7)); // ➞ 2
System.out.println(absDiff(6, -6)); // ➞ 12
System.out.println(absDiff(1.5, 1.5)); // ➞ 0
}
}
The time complexity of this solution is O(1) because it involves a constant number of operations regardless of the input values. The space complexity is also O(1) as no additional space is required.
Potential edge cases include:
absDiff(1.5, 1.5)
should return 0).absDiff(-5, -7)
should return 2).absDiff(0, 3)
should return 3).These edge cases are handled by the algorithm as it correctly computes the absolute difference in all scenarios.
To test the solution comprehensively, we should include a variety of test cases:
We can use the main method provided in the code implementation to run these test cases and verify the results.
When approaching such problems, it is essential to:
Practicing similar problems and studying different algorithms can help improve problem-solving skills.
In this blog post, we discussed how to find the absolute difference between two numbers in Java without using the built-in abs()
function. 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 programming.
For further reading and practice, consider the following resources: