1. Fizz Buzz.md

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

🚀 1. Fizz Buzz 🧠

The problem can be found at the following link: Problem Link

💡 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)

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)

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)

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. Let’s make this learning journey more collaborative!

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