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 C++ (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 non-negative 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 absolute difference 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:
// Naive solution to find the absolute difference
double absDiff(double a, double 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:
// Optimized solution using ternary operator
double absDiff(double a, double b) {
return (a < b) ? (b - a) : (a - b);
}
This approach is more compact and equally efficient.
Here is a step-by-step breakdown of the algorithm:
a
and b
.a
is less than b
, return b - a
.a - b
.The optimized solution uses the ternary operator to achieve the same logic in a single line of code.
// Function to find the absolute difference between two numbers
double absDiff(double a, double b) {
// Using ternary operator for a concise solution
return (a < b) ? (b - a) : (a - b);
}
This code is clean, readable, and follows best practices. The ternary operator makes the solution concise without sacrificing clarity.
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 beyond the input variables.
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 correctly by the provided solution.
To test the solution comprehensively, consider the following test cases:
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
These test cases cover a range of scenarios, including 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 abs()
function in C++. We explored a naive solution and an optimized solution, provided a detailed algorithm, and analyzed the complexity. We also covered edge cases and testing strategies.
Understanding and solving such problems is crucial for developing strong problem-solving skills. Practice regularly and explore further to enhance your coding abilities.
For further reading and practice, consider the following resources: