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Fizzbuzz

March 23, 2021 9:00 AM

I’ve seen several recent posts on the beginning-programmer forums about how to solve the fizzbuzz problem, so let’s talk about that today:

Enumerate the numbers from 1 to N by writing “Fizz” if the number is divisible by 3, “Buzz” if the number is divisible by 5, “FizzBuzz” if the number is divisible by both, and the number itself if the number is divisible by neither. For instance, counting from 1 to 25 works like this: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11. Fizz, 13, 14, FizzBuzz, 16, 17, Fizz, 19, Buzz, Fizz, 22, 23, Fizz, Buzz.

FizzBuzz also appears as the first problem in Project Euler, where they characterize the problem like this:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

Your task is to solve the two versions of the FizzBuzz problem, for all numbers up to N; you can use a simple method, but it is more fun to be a little bit clever. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Posted by programmingpraxis

Categories: Exercises

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8 Responses to “Fizzbuzz”

1. (defun fizz-buzz-sum (N)
     (loop for i from 1 below N
           when (or (zerop (mod i 3)) (zerop (mod i 5)))
             sum i))
   
   (mapcar 'fizz-buzz-sum '(10 1000))
   ; --> (23 233168)
   
   (defun fizz-buzz (N)
     (loop for i from 1 below N
           if (zerop (mod i 3))
             if (zerop (mod i 5))
               do (write-line "FizzBuzz")
             else
               do (write-line "Fizz")
             end
           else
             if (zerop (mod i 5))
               do (write-line "Buzz")
             else
               do (format t "~D~%" i)
             end
           end))
   
   (fizz-buzz 20)
   
   1
   2
   Fizz
   4
   Buzz
   Fizz
   7
   8
   Fizz
   Buzz
   11
   Fizz
   13
   14
   FizzBuzz
   16
   17
   Fizz
   19
   
   

   By Pascal Bourguignon on March 23, 2021 at 9:14 AM

2. There’s no reason to be clever, otherwise, you end up with something like: https://github.com/EnterpriseQualityCoding/FizzBuzzEnterpriseEdition ;-)

   By Pascal Bourguignon on March 23, 2021 at 9:15 AM

3. Here’s a solution in Python.

   N = 25
   
   for i in range(1, N + 1):
       s = ''
       if i % 3 == 0:
           s += 'Fizz'
       if i % 5 == 0:
           s += 'Buzz'
       if not s:
           s = str(i)
       print(s)
   
   print('----------')
   
   print(sum(x for x in range(1, 1000) if x % 3 == 0 or x % 5 == 0))
   

   Output:

   1
   2
   Fizz
   4
   Buzz
   Fizz
   7
   8
   Fizz
   Buzz
   11
   Fizz
   13
   14
   FizzBuzz
   16
   17
   Fizz
   19
   Buzz
   Fizz
   22
   23
   Fizz
   Buzz
   ----------
   233168
   

   By Daniel on March 23, 2021 at 4:44 PM

4. “you can use a simple method, but it is more fun to be a little bit clever”

   Here’s my attempt at a clever solution.

   for i in range(1, 26):
       s = 'Fizz' if not i % 3 else ''
       s += 'Buzz' if not i % 5 else ''
       print(s if s else i)
   

   By Daniel on March 23, 2021 at 4:51 PM

5. A solution in Racket:

   (define (fizzbuzz)
     (define dct #hash((0 . FizzBuzz) (6 . Fizz) (10 . Buzz)))
     (for ((n (in-range 1 101)))
       (let ((elr (modulo (expt n 4) 15)))
         (displayln (if (= 1 elr) n (dict-ref dct elr))))))
   

   By Kevin on March 23, 2021 at 6:03 PM

6. Another solution in Racket:
   (The first post was for 1 – 100, as requested by Project Euler; this post is for 1 to N.)

   (define (fizzbuzz N)
     (let ((dct #hash((0 . FizzBuzz) (6 . Fizz) (10 . Buzz))) (n-max (add1 N)))
       (for ((n (in-range 1 n-max)))
         (let ((elr (modulo (expt n 4) 15)))
           (displayln (if (= 1 elr) n (dict-ref dct elr)))))))
   

   Example:

   > (fizzbuzz 15)
   1
   2
   Fizz
   4
   Buzz
   Fizz
   7
   8
   Fizz
   Buzz
   11
   Fizz
   13
   14
   FizzBuzz
   

   By Kevin on March 23, 2021 at 6:30 PM

7. For the Sum from 1 to N in C#:

   (3(N/3)(N/3 + 1) + 5(N/5)(N/5 + 1) – 15(N/15)(N/15 + 1))/2

   By Ernest Gore on March 23, 2021 at 11:42 PM

8. my answer in c++

   #include
using namespace std;
int main()
{
int setc = 0;
int N = 100;
for (int i = 1; i < N; i++)
{
setc = 0;
if (i % 3 == 0)
{
setc = 1;
cout << "Fizz";
}
if (i % 5 == 0)
cout << "Buzz";

       else if(setc==0)
           cout << i;
       cout << endl;
   
   
   }
   return 0;
   

   }


   By faisi on April 6, 2021 at 4:41 PM

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