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# Fizz Buzz LeetCode Solution

## Problem – Fizz Buzz LeetCode Solution

Given an integer `n`, return a string array `answer` (1-indexed) where:

• `answer[i] == "FizzBuzz"` if `i` is divisible by `3` and `5`.
• `answer[i] == "Fizz"` if `i` is divisible by `3`.
• `answer[i] == "Buzz"` if `i` is divisible by `5`.
• `answer[i] == i` (as a string) if none of the above conditions are true.

Example 1:

``````Input: n = 3
Output: ["1","2","Fizz"]
``````

Example 2:

``````Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]``````

Example 3:

``````Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]``````

Constraints:

• `1 <= n <= 104`

### Fizz Buzz LeetCode Solution in Java

``````public class Solution {
public List<String> fizzBuzz(int n) {
List<String> ret = new ArrayList<String>(n);
for(int i=1,fizz=0,buzz=0;i<=n ;i++){
fizz++;
buzz++;
if(fizz==3 && buzz==5){
fizz=0;
buzz=0;
}else if(fizz==3){
fizz=0;
}else if(buzz==5){
buzz=0;
}else{
}
}
return ret;
}
}
``````

### Fizz Buzz LeetCode Solution in C++

``````class Solution {
public:
vector<string> fizzBuzz(int n) {
vector<string> res(n);
for(int i = 1;i <= n; i++) {
res[i - 1] = to_string(i);
}
for(int i = 2;i < n; i += 3) {
res[i] = "Fizz";
}
for(int i = 4;i < n; i += 5) {
res[i] = "Buzz";
}
for(int i = 14;i < n; i += 15) {
res[i] = "FizzBuzz";
}
return res;
}
};
``````

### Fizz Buzz LeetCode Solution in Python

``````def fizzBuzz(self, n):
return['FizzBuzz'[i%-3&-4:i%-5&8^12]or`i`for i in range(1,n+1)]
``````
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