My Solution for FizzBuzz (By A Technical Recruiter)


Not one to pass up my chance at a good programming problem, I took a stab at FizzBuzz because it seemed, well, relatively simply to construct a solution.

Here’s my code on gist:

First, I chose javascript because I could write it in notepad and run it on my PC (I don’t have coding tools like eclipse or vi installed here.  Why would I? I am a recruiter!)

Second, I just iterated through the problem with a foor loop 100 times and at each step, I did three tests:

  1. Modulus the integer by 3. If it’s equal to zero (meaning the integer divides exactly by three with no remainder) then add “fizz” to a variable, p, which was set to “” (blank) at the beginning of each set of tests.
  2. Modulus the integer by 5 (same gameshow as above, just round 2). If this is equal to zero, then I simply add “buzz” to the variable p. Bonus here is that p may already be equal to either blank or to “fizz” depending on the result of the previous test.  The challenge states that if the integer is a multiple of both three and five, write “fizzbuzz”. This accomplishes this by either appending “buzz” to “fizz” if they are both multiples, or simply setting p to “buzz” if only five is a multiple of the integer.
  3. Finally, if p is still equal to blank after these prior tests, just set it to the value of the integer.
  4. Last, write p to the page along with a “<br />” linebreak in html to bump the output down a line and make the whole thing a little more human readable.

Where I went wrong:

  • In my first implementation, i, the counter, was instantiated at 0 (zero), which is outside the bounds of the program guideline (and was a false-positive to both my modulus tests). That generated a false positive “fizzbuzz” as the first result AND set the entire set of results off by one integer.  I could have also solved this by using another variable in the code that was set to i+1 each time through the loop, but that seemed redundant and a waste compared to instantiating i at 1 instead of 0.
  • Second, I counted only up to 100, which left off the last integer to test, 100 itself, since I was starting at 1, not zero. I changed this by looping through until i<101. I could have also solved this by editing this part of the for loop to i<=100.

So, lets see your code!??

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Robert Merrill

Senior Technical Recruiter
I am a Husband, Dad, Recruitinator, Geek. Mormon. Smart-Alec. . Professionally, I am the Senior Manager of Talent Acquisition at", the world's largest online family history resource. Check out our jobs! Since 2001, I have helped connect companies and talented professionals with a particular style of recruiting, interviewing, onboarding and retention that helps companies change their recruiting processes and capabilities forever. I believe that even small companies can do fantastic recruiting with the right tools and mindset.

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