---
Difficulty: Easy
Source: 160 Days of Problem Solving (BONUS PROBLEMS 🎁)
Tags:
  - Strings
  - Modulus
  - Arrays
---

# 🚀 _1. Fizz Buzz_ 🧠

The problem can be found at the following link: [Problem Link](https://www.geeksforgeeks.org/batch/gfg-160-problems/track/string-bonus-problems/problem/fizz-buzz)

## 💡 **Problem Description:**

Given an integer `n`, for every integer `1 ≤ i ≤ n`, the task is to return an array of strings where:

- "FizzBuzz" is included if `i` is divisible by both 3 and 5,
- "Fizz" is included if `i` is divisible only by 3,
- "Buzz" is included if `i` is divisible only by 5,
- The number `i` itself is included as a string if none of the above conditions are true.

### **Examples:**

**Input:**

```
n = 3
```

**Output:**

```
["1", "2", "Fizz"]
```

**Input:**

```
n = 10
```

**Output:**

```
["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz"]
```

**Input:**

```
n = 20
```

**Output:**

```
["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz", "13", "14", "FizzBuzz", "16", "17", "Fizz", "19", "Buzz"]
```

### Constraints:

- $`1 ≤ n ≤ 10^6`$

## 🎯 **My Approach:**

### Step-by-Step:

1. **Iterate from 1 to `n`:**  
   Loop through all integers from 1 to `n`.

2. **Check divisibility:**

   - Use modulus (`%`) to check if `i` is divisible by both 3 and 5, only by 3, or only by 5.
   - If divisible by both 3 and 5, append `"FizzBuzz"`.
   - If divisible only by 3, append `"Fizz"`.
   - If divisible only by 5, append `"Buzz"`.
   - Otherwise, append the number itself as a string.

3. **Store Results:**  
   Add the computed value to the result list or array.

4. **Output Results:**  
   Return the result array.

## 🕒 **Time and Auxiliary Space Complexity**

- **Time Complexity:** O(n)  
  The algorithm iterates once through all integers from 1 to `n`. Each iteration involves constant-time operations to check divisibility and string concatenation.

- **Space Complexity:** O(n)  
  The result array stores all `n` strings.

## 📝 **Solution Code**

## Code (Cpp)

```cpp
class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> result;
        result.reserve(n);

        for (int i = 1; i <= n; i++) {
            string current = "";

            if (i % 3 == 0) current += "Fizz";
            if (i % 5 == 0) current += "Buzz";

            if (current.empty()) current = to_string(i);

            result.push_back(current);
        }

        return result;
    }
};
```

## Code (Java)

```java
class Solution {
    public static ArrayList<String> fizzBuzz(int n) {
        ArrayList<String> result = new ArrayList<>(n);

        for (int i = 1; i <= n; i++) {
            StringBuilder current = new StringBuilder();

            if (i % 3 == 0) current.append("Fizz");
            if (i % 5 == 0) current.append("Buzz");

            if (current.length() == 0) current.append(i);

            result.add(current.toString());
        }

        return result;
    }
}
```

## Code (Python)

```python
class Solution:
    def fizzBuzz(self, n: int):
        result = []

        for i in range(1, n + 1):
            current = []

            if i % 3 == 0:
                current.append("Fizz")
            if i % 5 == 0:
                current.append("Buzz")

            if not current:
                current.append(str(i))

            result.append("".join(current))

        return result
```

## 🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: [Any Questions](https://www.linkedin.com/in/patel-hetkumar-sandipbhai-8b110525a/). Let’s make this learning journey more collaborative!

⭐ If you find this helpful, please give this repository a star! ⭐

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