Fizzbuzz

March 23, 2021

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.

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Posted by programmingpraxis
Filed in Exercises
8 Comments »

8 Responses to “Fizzbuzz”

  1. Pascal Bourguignon said
    March 23, 2021 at 9:14 AM 
    (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
  2. Pascal Bourguignon said
    March 23, 2021 at 9:15 AM

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

  3. Daniel said
    March 23, 2021 at 4:44 PM

    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
  4. Daniel said
    March 23, 2021 at 4:51 PM

    “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)
  5. Kevin said
    March 23, 2021 at 6:03 PM

    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))))))
  6. Kevin said
    March 23, 2021 at 6:30 PM

    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
  7. Ernest Gore said
    March 23, 2021 at 11:42 PM

    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

  8. faisi said
    April 6, 2021 at 4:41 PM

    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;

    }

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