Self Explanatory Cpp Code !! 
IntuitionWe were supposed to return a vector, so okay we can easily make a vector of strings and return it with the content required.Approach
To check whether t
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We were supposed to return a vector, so okay we can easily make a vector of strings and return it with the content required. 

# Approach
<!-- Describe your approach to solving the problem. -->
1. To check whether the number is divisible by 3 or 5 or 3 and 5 both.
-> The term which I used `!(i%3)` or similar ones indicate here that if `i%3==0` and `!` of that, i.e., `!0` is `1` that is true.
2. If the above conditions don't hold good then we will simply add `i` as a string with the help of `toString()` function.

# Complexity
- Time complexity: Since a  `for() loop` which always goes from `1  to n` is there, so the  `TIME COMPLEXITY` herein becomes $$O(n)$$  
- Space complexity: A vector was made of size `n` so, $$O(n)$$.

**Just A Single Request**

![image.png](https://assets.leetcode.com/users/images/13848c18-f742-4399-80b7-ff19e6c00d3c_1738078115.1048074.png)
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# Code
```cpp []
class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> s(n);
        for(int i = 1; i <= n; i++) {
            if(!(i % 3) && !(i % 5)) {
                s.push_back("FizzBuzz");
            }
            else if(!(i % 3)) {
                s.push_back("Fizz");
            }
            else if(!(i % 5)) {
                s.push_back("Buzz");
            }
            else {
                s.push_back(to_string(i));
            }
        }
        return s;
    }
};

```
['C++']