[Community Puzzle] De-FizzBuzzer

Hey guys,

It’s my first time posting here, so let me know if I’m breaking any rules. (I did not find a topic dedicated to this puzzle so I created this one but don’t hesitate to point me to the right topic if it exists)

I am working on the De-FizzBuzzer puzzle and I validate test cases 1, 2 and 4. However I think I am misunderstanding something:

In one test case I have BuzzBuzzBuzz as input and the test case tells me that the expected value is 25, how can it be?

If I take 25 and applies the rules I get:

  • produce Fizz for each time the digit 3 appears in the integer => Don’t produce anything, there is no 3
  • produce Fizz for each time the integer can be divided by 3; =>Don’t produce anything 25%3 != 0
  • produce Buzz for each time the digit 5 appears in the integer; => Produce one Buzz as there is one 5 in 25
  • produce Buzz for each time the integer can be divided by 5; => Produce one Buzz as 25%5 == 0
  • if no Fizz or Buzz has been produced so far, produce the original integer as the substring. => We already produced something so don’t do that.

Which gives me the string BuzzBuzz for 25.

Could someone explain to me how 25 can produce BuzzBuzzBuzz following the rules of the puzzle?

Thank you in advance!

  • produce Buzz for each time the digit 5 appears in the integer;
  • produce Buzz for each time the integer can be divided by 5;

25 contains 5, thus Buzz
25 = 5*5, thus BuzzBuzz
Concatenate both and you have BuzzBuzzBuzz

Hi,
I don’t understand the logic.
For the first example number from 1 to 20 are :
1
2
FizzFizz
4
BuzzBuzz
Fizz
7
8
FizzFizz
Buzz
11
Fizz
Fizz
14
FizzBuzzBuzz
16
17
FizzFizz
19
Buzz

So if we add Fizz forEach multiple of 3 :
-> 6 should be FizzFizz and not only Fizz because 6 = 2x3

I have the same problem with BuzzBuzzBuzz, if found 55 and not 25

Thank for helping

6 (Fizz)
time the digit 3 appears: 0
time the integer can be divided by 3: 1 (6/3 = 2, 2 is not divisible by 3)

25 (BuzzBuzzBuzz)
time the digit 5 appears: 1
time the integer can be divided by 5: 2 (25/5 = 5, 5/5 = 1)

125 (BuzzBuzzBuzzBuzz)
time the digit 5 appears: 1
time the integer can be divided by 5: 3 (125/5 = 25, 25/5 = 5, 5/5 = 1)

2 Likes

Thank a lot for your explanation !
I understand know, my english is not helping :slight_smile:

1 Like

Hi,
all tests ok but validator 4,no.
I think it is an “ERROR” case I dont’t get.
I check if 3 or 5 is in the number and if any lettre not in(FizBu) are in the word.
Could please someone check with the validator ?
Thx

1 Like

There can be other characters than letters.