The core challenge of this problem is to determine whether a given non-negative integer n 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:
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 (%) to determine the remainder when n is divided by 2. If the remainder is 0, the number is even; otherwise, it is odd.
A naive solution would involve checking each number from 0 to n to see if it is even or odd. However, this is unnecessary and inefficient.
The optimized solution involves a single check using the modulo operator:
Here is a step-by-step breakdown of the algorithm:
// Function to check the parity of a number
function checkParity(n) {
// Check if the number is even
if (n % 2 === 0) {
return "even";
} else {
return "odd";
}
}
// Example usage
console.log(checkParity(12)); // Output: "even"
console.log(checkParity(13)); // Output: "odd"
The time complexity of this solution is O(1) because it involves a single modulo operation and a conditional check, both of which are constant-time operations.
The space complexity is also O(1) as no additional space is required.
Potential edge cases include:
Examples:
Input: n = 0 Output: "even" Input: n = 999999999 Output: "odd"
To test the solution comprehensively, consider the following test cases:
Use a testing framework like Jest or Mocha for automated testing.
When approaching such problems:
In this blog post, we discussed how to determine the parity of a number using a simple and efficient algorithm. Understanding and solving such problems is crucial for developing strong problem-solving skills in computer science.
We encourage readers to practice and explore further to deepen their understanding.