# Fizz Buzz Leetcode

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

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

Table of Contents

Input:

n=4

Output:

1

2

Fizz

4

## Algorithm for Fizz Buzz

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

References

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