fizz buzz python

Solutions on MaxInterview for fizz buzz python by the best coders in the world

showing results for - "fizz buzz python"
Adrián
12 Apr 2018
1def fizz_buzz(input):
2    if (input % 3 == 0) and (input % 5 == 0):
3        return "FizzBuzz"
4    if input % 3 == 0:
5        return "Fizz"
6    if input % 5 == 0:
7        return "Buzz"
8    else:
9        return input
10
11
12print(fizz_buzz(3))
Oskar
18 Nov 2019
1output = ""
2
3for i in range(1, 101):
4 
5    if (i % 3 == 0):
6        output += "Fizz"
7    
8    if (i % 5 == 0):
9        output += "Buzz"
10    
11    elif (i % 3 != 0): 
12        output += str(i)
13    
14    output += "\n" # Add a new line at the end of the output 
15
16print(output)   
Federico
24 Aug 2019
1#made by myself :)
2def FizzBuzz() :
3    Numbers = 0
4    for Numbers in range(101):
5        if Numbers%3 == 0:
6            print ("Fizz")
7        if Numbers%5 == 0:
8            print ("Buzz")
9        if Numbers%3 == 0 and Numbers%5 ==0:
10            print ("Fizz Buzz")
11        else :
12            print (Numbers)
13FizzBuzz()
Marta
28 Mar 2020
1fizz = 3
2buzz = 5
3fizzbuzz = fizz * buzz
4
5for i in range(100):
6  if i % fizzbuzz == o:
7    print('fizzbuzz')
8   elif i % fizz == 0:
9    print('fizz')
10   elif i % buzz == 0: 
11    print('buzz')
12   else:
13    print(i)
Juliana
13 Mar 2019
1def fizzBuzz(n):
2    # Write your code here
3    for i in range(1,a):
4        if(i%3==0 and i%5!=0):
5            print("Fizz")
6        elif(i%5==0 and i%3!=0):
7            print("Buzz")
8        elif(i%5==0 and i%3==0):
9            print(i)
10a = int(input("enter the no"))
11fizzBuzz(a)
12
Luce
10 Jul 2016
1# changeable game of fizz buzz (all you have to change is n and D
2# to add or alter the game.
3# No adding or changing if/else required to include bizz at 7 or fuzz at 11
4# just insert it into the dictionary. (you could easily make a function out of this)
5
6n = 100
7D = {3: 'fizz',
8     5: 'buzz'}
9for x in range(1, n+1):         # makes a range as big as we like
10    out = ''
11    for k, v in D.items():      # compares each in range with each key in dictionary
12        if x % k == 0:
13            out += v            # then attaches the associated word onto the output
14    if not out:
15        out = x                 # if the output is still empty, i.e. no listed factors
16    print(out, end=' ')         # the unchanged number becomes the output
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