[Practice Python Question] FizzBuzz | Python-3 Solution by APDaga

I came across a Python 3 practice question. 

Here, I am providing my solution to the problem "FizzBuzz" with the intention to help you learn and understand Python 3 in a better way.

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Python Problem Statement:

Given a number n, for each integer i in the range from 1 to n inclusive, print one value per line as follows:

• If i is a multiple of both 3 and 5, print FizzBuzz.
  
  
• If i is a multiple of 3 (but not 5), print Fizz.
  
  
• If i is a multiple of 5 (but not 3), print Buzz.
  
  
• If i is not a multiple of 3 or 5, print the value of i.
  

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Function Description:

Complete the function fizzBuzz in the editor below.

fizzBuzz has the following parameter(s):

• int n: upper limit of values to test (inclusive)
  
  
• Returns: NONE
  
  
• Prints:The function must print the appropriate response for each value i in the set {1, 2, ... n} in ascending order, each on a separate line.

Constraints:

0 < n < 2 × 10^5

Sample Input:

STDIN    Function
-----    --------
15    →  n = 15

Sample Output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

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Explanation:

• The numbers 3, 6, 9, and 12 are multiples of 3 (but not 5), so print Fizz on those lines.
  
  
• The numbers 5 and 10 are multiples of 5 (but not 3), so print Buzz on those lines.
  
  
• The number 15 is a multiple of both 3 and 5, so print FizzBuzz on that line.
  
  
• None of the other values is a multiple of either 3 or 5, so print the value of i on those lines.

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Python 3 Solution:

#!/bin/python3 ...
#
# Complete the 'fizzBuzz' function below.
# The function accepts INTEGER n as parameter.

def fizzBuzz(n):
	# Write your code here
	i=1;
	while i<=n:
		if i%3==0 and i%5==0:
			print('FizzBuzz')
	        elif i%3==0 and i%5!=0:
		        print('Fizz')
	        elif i%3!=0 and i%5==0:
		        print('Buzz')
	        elif i%3!=0 and i%5!=0:
		        print(i)
	        i+=1

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-Akshay P Daga