Home/ Questions/Q 14147
In Process

askthedev.com Latest Questions

Asked: 2024-09-27T01:18:48+05:30

How to implement the 1-2-Fizz-4-Buzz coding challenge in various programming languages?

I stumbled upon this fun coding challenge the other day that really got my brain buzzing, and I thought it would be interesting to share it here and see how others tackle it. The challenge is a twist on the classic “FizzBuzz” game we’ve all heard of (or participated in), and it’s called “1-2-Fizz-4-Buzz”.

So here’s the gist: instead of the usual counting up from 1, you have to print numbers from 1 to N, where N is provided as input. But there are a few rules that change how you print those numbers:

1. For every number that is divisible by 3, you should print “Fizz” instead of the number.
2. For every number that is divisible by 5, you should print “Buzz”.
3. If a number is divisible by both 3 and 5, you print “FizzBuzz”.
4. Now here’s where it gets a little more interesting! If a number is divisible by 2, you switch it up and instead of just printing the number or Fizz/Buzz, you print a list where “1” is replaced by “Fizz,” “2” keeps its number, and so on, basically alternating between numbers and the Fizz rules you’ve set.

It sounds pretty straightforward, but here’s the catch: when you hit a number that’s divisible by both 2 and 3 or 5, you need to sprinkle both sets of rules onto that number. It’s like a layered approach to the output.

For example, if you were to run this with a small value of N, like 10, you should expect an output that starts looking something like this: “1”, “Fizz”, “Fizz Buzz”, “4”, “Fizz”, “Buzz”, “Fizz”, and so forth.

I guess my question for you all is: what’s your strategy for implementing this in your preferred programming language? Any tips on keeping your code concise while still being readable? Or do you prefer going all out with complex nested conditions? Looking forward to hearing your thoughts and maybe seeing some fun snippets! Would love to see how different minds tackle the same problem.

Coding Challenge
Share

    Leave an answer

    You must login to add an answer.

    Need An Account,

    2 Answers

    1. function fizzBuzz(n) {
    for (let i = 1; i <= n; i++) {
        let output = '';
        
        // Check for FizzBuzz first
        if (i % 3 === 0 && i % 5 === 0) {
            output = 'FizzBuzz';
        } 
        // Check for Fizz
        else if (i % 3 === 0) {
            output = 'Fizz';
        } 
        // Check for Buzz
        else if (i % 5 === 0) {
            output = 'Buzz';
        }
        
        // If not Fizz or Buzz, check for even numbers
        if (i % 2 === 0) {
            if (output) {
                // If there's already output, concatenate
                output += ' ';
            }
            // Create the alternate list for the even numbers
            let altOutput = '';
            for (let j = 1; j <= i; j++) {
                if (j % 3 === 0 && j % 5 === 0) {
                    altOutput += 'FizzBuzz ';
                } else if (j % 3 === 0) {
                    altOutput += 'Fizz ';
                } else if (j % 5 === 0) {
                    altOutput += 'Buzz ';
                } else {
                    altOutput += j + ' ';
                }
            }
            output += altOutput.trim(); // Remove trailing space
        }
        
        // If nothing matched, just output the number
        if (!output) {
            output = i;
        }
        
        console.log(output);
    }
}

// Run it with a value for N, e.g., 10
fizzBuzz(10);

    2. This “1-2-Fizz-4-Buzz” challenge is a creative twist on the classic FizzBuzz and can be tackled efficiently with a straightforward approach in code. A clean solution would involve iterating through the numbers from 1 to N and using a series of conditional statements to handle the output criteria. Here’s a Python implementation that maintains readability while following the rules you’ve provided:

      def fizz_buzz_n(n):
    for i in range(1, n + 1):
        output = ''
        if i % 3 == 0:
            output += 'Fizz'
        if i % 5 == 0:
            output += 'Buzz'
        if i % 2 == 0:
            if output:
                output += ' '  # Only add space if there's existing output
            fizz_list = ['Fizz' if j % 3 == 0 else str(j) for j in range(1, i + 1)]
            output += ', '.join(fizz_list)
        else:
            output = output or str(i)
        print(output)

# Example usage:
fizz_buzz_n(10)

      The function `fizz_buzz_n` executes a loop from 1 to N and builds the output string based on the divisibility conditions. By combining the core rules for Fizz and Buzz with the additional logic for even numbers, this approach keeps the code structured and easy to follow. For those keen on keeping their code concise, list comprehensions can promote clarity, while maintaining functionality for generating the Fizz output. This method helps avoid complex nested conditions while delivering the results as specified.

    Related Questions