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 JavaScript (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 Math.abs()
function. The absolute difference is always a positive number, regardless of the order of the input numbers.
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 a simple if
statement.
A naive solution would involve checking the condition and returning the appropriate difference:
function absDiff(a, b) {
if (a < b) {
return b - a;
} else {
return a - b;
}
}
This solution is straightforward and works correctly, but it can be optimized further.
An optimized solution can use the ternary operator to make the code more concise:
function absDiff(a, b) {
return a < b ? b - a : a - b;
}
This approach is more compact and achieves the same result.
Let's break down the algorithm step-by-step:
a
and b
.a
is less than b
, return b - a
.a - b
.Here is the well-commented code implementation in JavaScript:
/**
* Function to calculate the absolute difference between two numbers
* @param {number} a - First number
* @param {number} b - Second number
* @returns {number} - The absolute difference between a and b
*/
function absDiff(a, b) {
// Check if a is less than b
if (a < b) {
// Return the difference b - a
return b - a;
} else {
// Return the difference a - b
return a - b;
}
}
// Optimized version using ternary operator
function absDiffOptimized(a, b) {
// Return the absolute difference using ternary operator
return a < b ? b - a : a - b;
}
The time complexity of both the naive and optimized solutions is O(1) because the operations involved (comparison and subtraction) take constant time.
The space complexity is also O(1) as no additional space is required beyond the input variables.
Consider the following edge cases:
absDiff(1.5, 1.5) ➞ 0
absDiff(-5, -7) ➞ 2
absDiff(0, 3) ➞ 3
Each algorithm handles these edge cases correctly by ensuring the result is always positive.
To test the solution comprehensively, consider a variety of test cases:
console.log(absDiff(3, 0)); // ➞ 3
console.log(absDiff(0, 3)); // ➞ 3
console.log(absDiff(10, -1)); // ➞ 11
console.log(absDiff(-5, -7)); // ➞ 2
console.log(absDiff(6, -6)); // ➞ 12
console.log(absDiff(1.5, 1.5)); // ➞ 0
These test cases cover positive numbers, negative numbers, and zero.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the absolute difference between two numbers without using the built-in Math.abs()
function. We explored a naive solution and an optimized solution, analyzed their complexities, and considered edge cases. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
Keep practicing and exploring further to enhance your coding abilities!
For further reading and practice problems, consider the following resources: