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### LeetCode Solution - Problem Fizz Buzz

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`
Solution:

Java:

``````class Solution {
public List<String> fizzBuzz(int n) {
List<String> list = new ArrayList<>();
for (int i=1; i <= n; i++) {
if (i % 3 == 0 && i % 5 == 0) {
list.add("FizzBuzz");
} else if (i % 3 == 0) {
list.add("Fizz");
} else if ( i % 5 == 0) {
list.add("Buzz");
} else {
list.add(Integer.toString(i));
}
}
return list;
}
}
``````
Python3:

``````class Solution:
def fizzBuzz(self, n: int) -> List[str]:
ans = []
for i in range(1,n+1):
if i%3 == 0 and i%5 == 0:
ans.append("FizzBuzz")
elif i%3 == 0:
ans.append("Fizz")
elif i%5 == 0:
ans.append("Buzz")
else:
ans.append(str(i))
return ans
``````

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