Fizz buzz on python

Python FizzBuzz [closed]

The best part is how you remembered elif only after writing else if , even though the problem statement told you to remember it.

There are many Python tutorials that will help you to get a basic understanding of the language. For example docs.python.org/2/tutorial is the official introduction.

11 Answers 11

A few issues with your code here. The first issue is that, for comparison, you should be using == , not = , which is for assignment.

The second issue is that you want to check that the remainder of the divisions (which is what the modulo operator calculates) is zero, not that it’s true, which doesn’t really make sense.

You should be using elif for «otherwise if. » and else for «otherwise.» And you need to fix the formatting of your else clause.

n=input() if n%3 == 0: print("Fizz") elif n%5 == 0: print ("Buzz") else: print n 

Finally, your code does not meet the spec:

1) If the number is divisible by both 3 and 5 print «FizzBuzz»

The above will not do this. This part I’m going to leave to you because I’m not here to solve the assignment for you 🙂

Also, since the OP is using print n , I’m assuming he’s on 2.x where True == 1 and False == 0 . Not a great way to write the code, but nothing wrong with it.

FizzBuzz: For integers up to and including 100, prints FizzBuzz if the integer is divisible by 3 and 5 (15); Fizz if it’s divisible by 3 (and not 5); Buzz if it’s divisible by 5 (and not 3); and the integer otherwise.

def FizzBuzz(): for i in range(1,101): print < 3 : "Fizz", 5 : "Buzz", 15 : "FizzBuzz">.get(15*(not i%15) or 5*(not i%5 ) or 3*(not i%3 ), '<>'.format(i)) 

The .get() method works wonders here.

For all integers from 1 to 100 (101 is NOT included),
print the value of the dictionary key that we call via get according to these rules.

«Get the first non-False item in the get call, or return the integer as a string.»

When checking for a True value, thus a value we can lookup, Python evaluates 0 to False . If i mod 15 = 0, that’s False, we would go to the next one.

Therefore we NOT each of the ‘mods’ (aka remainder), so that if the mod == 0, which == False, we get a True statement. We multiply True by the dictionary key which returns the dictionary key (i.e. 3*True == 3 )

When the integer it not divisible by 3, 5 or 15, then we fall to the default clause of printing the int ‘<>‘.format(i) just inserts i into that string — as a string.

Fizz
79
Buzz
Fizz
82
83
Fizz
Buzz
86
Fizz
88
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz

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Fizz Buzz in Python

Welcome to Fizz Buzz in Every Language! In this series, we’ll be implementing Fizz Buzz in as many languages as possible. Up first, let’s implement Fizz Buzz in Python. Today is bit special because we have an article written by someone in the community, samdoj. Don’t forget to thank them for their contribution in the comments! If you don’t know anything about Python, we recommend checking out Hello World in Python. At any rate, let’s dive in!

Table of Contents

Fizz Buzz in Python

for i in range(1, 101): line = '' if i % 3 == 0: line += "Fizz" if i % 5 == 0: line += "Buzz" if not line: line += str(i) print(line)

If a number is divisible by 3, print the word ‘Fizz’ instead of the number. If the number is divisible by 5, print the word ‘Buzz’ instead of the number. Finally, if the number is divisible by both 3 and 5, print ‘FizzBuzz’ instead of the number. Otherwise, just print the number.

You can test for divisibility using the modulo operator. The modulo operator divides two numbers and yields the remainder, so i modulo j is 0 if i is divisible by j . In Python, this is written as i % j . Then, it’s a simple matter of checking whether i % 3 == 0 or i % 5 == 0 .

Code Style

You’ll notice first how everything is properly indented. This is not just good code style, Python actually enforces it. There’s no need to declare variables as Python is what’s called a weakly typed language. That means it can figure out what type a variable should be on the fly.

The Loop

Control Flow

line = '' if i % 3 == 0: line += "Fizz" if i % 5 == 0: line += "Buzz" if not line: line += str(i)

If the number is divisible by 3, as explained above, we add the word “Fizz” to the empty string. If it’s divisible by 5, we add the word “Buzz”. Notice the efficiency here. We don’t need and because by simply adding “Buzz”, we meet the requirement for the case where the number is divisible by 3 and 5, or just 5. Then we add i to the empty string if the string is still empty. Notice that an empty string returns false. This is a concept called falsey. In a weakly typed language, like Python and JavaScript, values such as 0 , undefined , null , and » all return false when they’re used in logical comparisons.

Printing

Since we declare an empty string at every iteration, we don’t have to worry about line containing any junk from the previous iteration.

How to Run the Solution

To run the Fizz Buzz in Python program, grab a copy of the Python file from GitHub . After that, get the latest version of Python . Now, all you have to do is run the following from the command line:

Alternatively, you can always copy the source code into an online Python interpreter and hit run.

Sample Programs in Every Language

And, there you have it! We’ve successfully written a program to perform the Fizz Buzz algorithm in Python. If you liked this article, don’t forget to give it a share. Also, remember that you can contribute to this series by dropping your suggestions in the comments or forking the GitHub repository.

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Exciting FizzBuzz Challenge in Python With Solution

There are thousands of python learning platform where you can practice your Python coding skills. These platforms contain some of the best problems which you can ever imagine. The programs are separated into several categories depending on their topic category and difficulty level. These platforms definitely help you learn new things and improve your coding practices. In this post, we’ll go through the solutions of FizzBuzz Python.

FizzBuzz Python is a popular python question in HackerRank and HackerEarth learning platforms. Both the platforms have the same problem statement and are very special for new programmers. The program asks you to print “Fizz” for the multiple of 3, “Buzz” for the multiple of 5, and “FizzBuzz” for the multiple of both. In both the platforms, the best optimal solution for the program is expected, which takes the lowest time to execute.

In this post, we’ll go through all of the solutions in all languages, including python 2 and python 3.

What exactly is the FizzBuzz Python Problem Statement?

The exact wordings of the problem goes as –

Print every number from 1 to 100 (both included) on a new line. Numbers which are multiple of 3, print “Fizz” instead of a number. For the numbers which are multiples of 5, print “Buzz” instead of a number. For the number which is multiple of both 3 and 5, print “FizzBuzz” instead of numbers.

Problem statement seems very easy for an everyday programmer. But from a newbie’s perspective, this program tests the skills regarding loops and conditionals. Let’s have a look at the constraints given for the answers to be acceptable.

Constraints for the FizzBuzz Problem

Constraints are the limiting factors within which your code must comply. These constraints are made to identify better codes with minimum time complexity and better memory management. Following are the constraints for the FizzBuzz Python problem –

1. Time Limit: 5 seconds
2. Memory Limit: 256 MB
3. Source Limit: 1024KB
4. Scoring System: (200 – number of characters in source code)/100 [Only for python solutions]

Hints For FizzBuzz Python Problem

There are multiple ways to solve the FizzBuzz Python problem. If you want hints for the same here, they are –

Hint 1: Create a “for” loop with range() function to create a loop of all numbers from 1 to 100. Before implementing FizzBuzz, create this simple loop to understand the looping.

Hint 2: To check the number is a multiple of any number, check the remainder of the number with the divisor. If the remainder turns out to be 0, then it’s multiple of the corresponding number. For example, 15 leaves remainder 0 when divided by 5. This confirms that 15 is a multiple of 5. Use the same logic to create a logical conditional.

Hint 3: In conditional statements, put the multiple of 15 cases on top of 5 or 3. Because if the number is a multiple of 15, it’ll always be a multiple of 3 and 5. Implementing this will check for the FizzBuzz case first.

FizzBuzz Python 3 Solution

Solution for FizzBuzz problem in Python 3 –

for num in range(1, 101): if num % 15 == 0: print("FizzBuzz") elif num % 3 == 0: print("Fizz") elif num % 5 == 0: print("Buzz") else: print(num)

Explanation –

Firstly, we declare a loop that ranges from 1 to 100. As the range() function loops till inclusive integer, we’ve used 101. We’ve used the if statements from the next block to check if the multiplicity of every number. If it is divisible by 15, print “FizzBuzz,” if it’s divisible by 3, print “Fizz” if it’s divisible by 5, print “Buzz.” All these conditionals are combined by using if and elif blocks. This looping goes on until it reaches 100.

FizzBuzz Python 2 Solution

Solution for FizzBuzz problem in Python 2 –

for num in range(1, 101): if num % 15 == 0: print "FizzBuzz" elif num % 3 == 0: print "Fizz" elif num % 5 == 0: print "Buzz" else: print num

Explanation follows the same for python 2. The only difference being that the print function works without parenthesis.

Most Efficient Fizzbuzz Python

When it comes to solving python programs, the most efficient solution is best. Even if your code is long, it has to be efficient to compute less and give the same out. Here we have most efficient solution for Fizzbuzz which’ll help you to develop your algorithmic side of the brain.

for i in range(1,101): print([i,"buzz","fizz","fizzbuzz"][2*(i%3==0) + (i%5==0)])

Explanation:

There are several ways of completing the FizzBuzz problem. Each conditional statement takes O(1) time complexity. So having fewer conditional statements and creating a code that prevents large multiplications are best.

In our code, we’ve created a list that has a different output that is accessed by the indexing. As i%3==0 returns 1 if i is divisor or 3 and the same goes for i%5==0. By combining then with 2*(i%3==0) + (i%5==0) you can get a proper index of the list.

Fizzbuzz Python One Liner Solution

for i in range(1, 101): print("Fizz"*(i%3==0)+"Buzz"*(i%5==0) or str(i))

Explanation:

Python supports one-liner for loops included with conditional statements. FizzBuzz is a perfect problem where you can code the entire solution in one line. Using loops and conditionals in one line, you can score maximum points.

Solutions for FizzBuzz in Other Languages

Solving FizzBuzz Problem In C++

#include using namespace std; int main() < for(int i=1;i<=100;i++)< if((i%3 == 0) && (i%5==0)) cout<<"FizzBuzz\n"; else if(i%3 == 0) cout<<"Fizz\n"; else if(i%5 == 0) cout<<"Buzz\n"; else cout<return 0; >

Solving FizzBuzz Problem in Java 8

import java.io.*; import java.util.*; public class Solution < public static void main(String[] args) < int x = 100; for(int i = 1; i else if(i % 5 == 0) < System.out.println("Buzz"); >else if(i % 3 ==0) < System.out.println("Fizz"); >else < System.out.println(i); >> > >

FizzBuzz Problem In Go

package main import "fmt" func main() < for i := 1; i else if i%3 == 0 < fmt.Printf("Fizz\n") >else if i%5 == 0 < fmt.Printf("Buzz\n") >else < fmt.Printf("%d\n", i) >> >

Solving FizzBuzz Problem In Javascript (NodeJS v10)

process.stdin.resume(); process.stdin.setEncoding("utf-8"); var stdin_input = ""; process.stdin.on("data", function (input) < stdin_input += input; // Reading input from STDIN >); process.stdin.on("end", function () < main(stdin_input); >); function main(input) < var str; var i=1; while(i<=input)< str=''; if(i%3===0)< str+='Fizz'; >if(i%5===0) < str+='Buzz'; >str!=='' ? process.stdout.write(str+"\n") : process.stdout.write(i+"\n"); i++; > >

Solving FizzBuzz Problem In PHP

R (RScript 3.4.0)

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Conclusion

There are thousands of awesome problems that test your basic knowledge in the world of coding. These problems not only help you to learn to code but also improves your logical thinking. Hence, you should always practice coding problems even if you are in a job. There is no harm in learning more everything. To summarize, the FizzBuzz problem tests your basic coding knowledge.

Enjoy Learning and Enjoy Coding!

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