**FizzBuzz - Amazon Top Interview Questions**

### Problem Statement :

Given an integer n, return a list of integers from 1 to n as strings except for multiples of 3 use “Fizz” instead of the integer and for the multiples of 5 use “Buzz”. For integers which are multiples of both 3 and 5 use “FizzBuzz”. Constraints 0 ≤ n ≤ 100,000 Example 1 Input n = 15 Output ["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz", "13", "14", "FizzBuzz"]

### Solution :

` ````
Solution in C++ :
vector<string> solve(int n) {
vector<string> res;
for (int i = 1; i <= n; i++) {
string s;
if (i % 3 == 0) s += "Fizz";
if (i % 5 == 0) s += "Buzz";
res.push_back(s.empty() ? to_string(i) : s);
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public String[] solve(int n) {
String s[] = new String[n];
s[0] = "1";
for (int i = 1; i < s.length; i++) {
if ((i + 1) % 3 == 0 && (i + 1) % 5 == 0) {
s[i] = "FizzBuzz";
} else if ((i + 1) % 3 == 0) {
s[i] = "Fizz";
} else if ((i + 1) % 5 == 0) {
s[i] = "Buzz";
} else {
s[i] = String.valueOf(i + 1);
}
}
return s;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, n):
# return [self.convert_int(i) for i in range(1, n + 1)]
return [
"FizzBuzz" if not i % 15 else "Buzz" if not i % 5 else "Fizz" if not i % 3 else str(i)
for i in range(1, n + 1)
]
```

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