Day1

Day 1 -- The Fizz Buzz Challenge

Brief History - Fizz Buzz is a group word game for children to teach them about division. Players take turns to count incrementally, replacing any number divisible by three with the word "fizz", and any number divisible by five with the word "buzz" and any number divisible by both three and five is replaced with the word "fizzbuzz"

For example, a typical round of fizz buzz would start as follows: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, Fizz Buzz, 16, 17, Fizz, 19, Buzz, Fizz, 22, 23, Fizz, Buzz, 26, Fizz, 28, 29, Fizz Buzz, 31, 32, Fizz, 34, Buzz, Fizz, …

Question- Write a program that prints the numbers from 1 to n and for multiples of '3' print "Fizz" instead of the number, for the multiples of '5' print "Buzz", and for the numbers which are divisible by both 3 and 5, print FizzBuzz.

JavaScript Implementation

Solution 1

/**
 * @author MadhavBahlMD
 * @date 20/12/2018
 */

 // Method 1 (The general approach)
// Step 1: Run a loop from 1 to n, for reach iteration (i) perform the next steps
// Step 2: if i is divisible by both 3 and 5, print “FizzBuzz”
// Step 3: if above condition fails, and i is divisible by 3, print “Fizz”
// Step 4: if above condition fails, and i is divisible by 5, print “Buzz”
// Step 5: if all conditions fails, print i itself.

function fizzbuzz (num) {
    for (let i=1; i<=num; i++) {
        if (i%3 === 0 && i%5 === 0)
            console.log ("FizzBuzz");
        else if (i%3 === 0)
            console.log ("Fizz");
        else if (i%5 === 0)
            console.log ("Buzz");
        else
            console.log (i);
    }
}

fizzbuzz (17);

Solution 2

/**
 * @author MadhavBahlMD
 * @date 20/12/2018
 */

// Step 1: Run a loop from 1 to n, for reach iteration (i) perform the next steps
// Step 2: Declare a temporary empty string (inside loop, say output)
// Step 3: If i is divisible by 3, append Fizz to the output.
// Step 4: If i is divisible by 5, append Buzz to the output.
// Step 5: If output is still an empty string, set it equal to i
// Step 6: Print output.

function fizzbuzz (num) {
    for (let i=1; i<=num; i++) {
        let opStr = '';

        if (i%3 === 0)  opStr += 'Fizz';
        if (i%5 === 0)  opStr += 'Buzz';

        if (opStr === '')  opStr = i;

        console.log (opStr);
    }
}

fizzbuzz (17);

Java Implementation

Solution 1

/**
 * @author: MadhavBahlMD
 * @date: 20/12/2018
 */

import java.util.Scanner;

public class Fizzbuzz {
    public static void main(String[] args) {

        System.out.println("/* ===== Fizz Buzz ===== */");
        Scanner input = new Scanner(System.in);
        System.out.print("\nEnter a number: ");
        int n = input.nextInt();
        for (int i=1; i<=n; i++) {
            if (i%3 == 0 && i%5 == 0) {
                System.out.println("FizzBuzz");
            } else if (i%3 == 0) {
                System.out.println("Fizz");
            } else if (i%5 == 0) {
                System.out.println("Buzz");
            } else {
                System.out.println(i);
            }
        }

    }
}

C Implementation

Fizzbuzz.c

/**
 * @author: Rajdeep Roy Chowdhury<rrajdeeproychowdhury@gmail.com>
 * @github: https://github.com/razdeep
 * @date: 20/12/2018
**/

#include <stdio.h>

int main()
{
    int n;
    printf("Enter a number upto which you want to find Fizzbuzz numbers ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        if (i % 5 == 0 && i % 3 == 0)
        {
            printf("FizzBuzz\n");
        }
        else if (i % 5 == 0)
        {
            printf("Buzz\n");
        }
        else if (i % 3 == 0)
        {
            printf("Fizz\n");
        }
        else
        {
            printf("%d\n", i);
        }
    }
    return 0;
}

Python(3) Implementation

fizzbuzz.py

'''
 * @author: Hrishi Patel <hrishipatel99@gmail.com>
 * @github: https://github.com/hrishi1999
 * @date: 20/12/2018
'''

n = int(input("Enter number of digits: "))

for i in range(1, n):
  if i%3==0 and i%5==0:
    print("fizzbuzz")
  elif i%3==0:
    print("fizz")
  elif i%5==0:
    print("buzz")
  else:
    print(i)

fizzbuzz2.py

'''
 * @author: ashwek
 * @date: 20/12/2018
'''

n = int(input("Enter limit : "))

for i in range(1, n+1):
    
    output = ""
    if i%3 == 0 :
        output = "Fizz"
    if i%5 == 0:
        output += "Buzz"

    if output :
        print(output)
    else:
        print(i)

profgrammer_fizzbuzz.py

'''
  @author: profgrammer 
  @date: 30-12-2018
'''

n = int(input())

for i in range(1, n+1):
  if(i % 15 == 0):
    print("Fizz Buzz")
  elif(i % 5 == 0):
    print("Buzz")
  elif(i % 3 == 0):
    print("Fizz")
  else:
    print(i)

C++ Implementation

FizzBuzz.cpp

'''
 * @author: Deepak Sharma
 * @github: https://github.com/dsdsharma
 * @date: 20/12/2018
'''
#include <iostream>

using namespace std;

int main()
{
   int n;
   cout << "====Fizz Buzz ====" << endl;
   cout << "Enter the number: ";
   cin >> n;
   for(int i=1; i<=n; i++){
       if(i%5==0 && i%3==0){
           cout << "FizzBuzz ";
       }else if(i%3==0){
           cout << "Fizz ";
       }else if(i%5==0){
           cout << "Buzz ";
       }else{
           cout << i << " ";
       }

   }

   return 0;
}

C# Solution

/**
 * @author imkaka
 * @date 20/12/2018
 */


using System;

public class FizzBuzz{

    public static void Main(string[] args){

        Console.WriteLine("/*====== Fizz Buzz ======*/");

        Console.WriteLine("Enter a Number : ");
        int input = Convert.ToInt32(Console.ReadLine());

        for(int i=1; i <= input; ++i){

            string res = "";

            if(i % 3 == 0) res += "Fizz";
            if(i % 5 == 0) res += "Buzz";

            if(res == "") res = i.ToString();

            Console.WriteLine(res);
        }
    }
}

Go Implementation

FizzBuzz.go

/* 
  * @author: Ramanpreet Singh <ramanpreet6262@gmail.com> 
  * @github: https://github.com/Ramanpreet6262
  * @date: 20/12/2018
*/


package main

import "fmt"

func main() {
    fmt.Println("Enter a number upto which you want to find Fizzbuzz numbers: ")
    var n int
    _, err := fmt.Scanf("%d", &n)
    
    if err != nil {
            fmt.Println(err)
         }

    
    for i := 1; i <= n; i++ {
        if i % 3 == 0 && i % 5 == 0 {
            fmt.Println("FizzBuzz")
        } else if i % 5 == 0 {
            fmt.Println("Buzz")
        } else if i % 3 == 0 {
            fmt.Println("Fizz")
        } else {
            fmt.Println(i)
        }
    }
}

Ruby Implementation

Fizzbuzz.rb

=begin
 @author: aaditkamat
 @date: 21/12/2018
=end

def get_result(num)
  result = ''
  if num % 3 != 0 and num % 5 != 0
    result += num.to_s
  end
  if num % 3 == 0
    result += 'Fizz'
  end
  if num % 5 == 0
    result += 'Buzz'
  end
  result
end

def solution(n)
  n.times do |num|
  p get_result(num + 1)
  end
end

Python Implementation

ShashankFizz.py (./python/ShashankFizz.py)

"""
        * @author: Shashankjain12
        * @date: 21/12/18
"""
print("Fizz Buzz")
n=int(input("Enter the number?"));
for i in range(1,n+1):
    if((i%3)==0 and (i%5)==0):
        print("Fizzbuzz ")
    elif((i%3)==0):
        print("fizz ")
    elif ((i%5)==0):
        print("buzz ")
    else:
        print(i)

PHP Implementation

fizzbuzz.php (./php/fizzbuzz.php)

/*
 * @author willianwt
 * @date 16/01/2018
 */

$num = n; #any number will work
for ($i=1; $i <= $num; $i++) { 
  if ($i % 3 == 0 && $i % 5 == 0) {
    echo "FizzBuzz\n";
  } else if ($i % 3 == 0) {
   echo "Fizz\n";
  }else if ($i % 5 == 0) {
   echo "Buzz\n";
  }else{
    echo $i . "\n";
  }
  
}

Have Another solution?

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