In Fizz Buzz problem we have given a number n, print the string representation of numbers from 1 to n with the given conditions:

- Print “Fizz” for multiples of 3.
- Print “Buzz” for multiples of 5.
- Print “FizzBuzz” for multiples of both 3 and 5.
- Otherwise, print the number in string format.

## Example

Input:

n=4

Output:

1

2

Fizz

4

## Algorithm for Fizz Buzz

- Iterate on the numbers from 1 to n( loop variable is i).
- For every number, if it is divisible by both 3 and 5 i.e.,
**i%3=0 and i%5=0**, then print “FizzBuzz”. - Else, if the number is divisible by 3 i.e.,
**i%3=0**, then print “Fizz”. - Else, if the number is divisible by 5 i.e.,
**i%5=0**, print “Buzz”. - Else, print the number as a string.

## Implementation

### C++ Program for Fizz Buzz Leetcode

#include<bits/stdc++.h> using namespace std; void fizzbuzz(int n){ for(int i=1;i<=n;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<<to_string(i)<<"\n"; } } } int main(){ int n; cin>>n; fizzbuzz(n); }

14

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

### JAVA Program for Fizz Buzz Leetcode

import java.util.Scanner; class Main { static void fizzbuzz(int n){ 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(Integer.toString(i)); } } } public static void main(String args[]) { int n; Scanner sc = new Scanner(System.in); n = sc.nextInt(); fizzbuzz(n); } }

7

1 2 Fizz 4 Buzz Fizz 7

## Variations

Write “Fizz” for the number divisible by 4 and “Buzz” for number divisible by 8, print any valid answer. In this case, numbers that are divisible by 8 are also divisible by 4 as 8 is a multiple of 4. Hence there may be multiple valid answers for this question as on multiple of 8 we can either choose “Fizz” or “Buzz”. Even if you observe carefully then we can replace all the numbers which divisible by 4 with “Fizz”, that will also be a valid answer.

Note: Carefully observe while solving this question as we saw that if the given numbers are multiple of each other then we can have multiple valid answers.

## Complexity Analysis

**Time complexity**

**O(n) **where n is the number given to us. As we need to traverse every number once from 1 to N for print its a string format.

### Space Complexity

**O(1)** because we don’t use or create any extra auxiliary space.