📘 Day 2 – Teacher's Day Challenge | FizzBuzz Logic & Conditionals

🗓️ Date: July 26, 2025
🔢 Problems Solved Today: 1
🎯 Focus Topic: Loops, Conditional Logic, Strings

💻 Platform: Leetcode

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🔗 Table of Contents

• Problem 1 – FizzBuzz

• Daily Summary

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🧩 Problem 1 – FizzBuzz

🔗 LeetCode Link – FizzBuzz #412
📚 Difficulty: Easy
🧠 Concepts: Modulo, Conditional Logic, String Arrays

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📄 Problem Statement (Summary):
Given an integer n, return a string array (1-indexed) such that:

• "FizzBuzz" if divisible by both 3 and 5

• "Fizz" if divisible by 3

• "Buzz" if divisible by 5

• Else, the number itself as a string

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💡 Approach:

• Initialize a vector of size n

• Loop from 1 to n

• Check divisibility by 3 and 5 using %

• Assign appropriate string at index i-1 (1-indexed logic)

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🧪 Code (C++):

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> result(n);
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 == 0)
                result[i - 1] = "FizzBuzz";
            else if (i % 3 == 0)
                result[i - 1] = "Fizz";
            else if (i % 5 == 0)
                result[i - 1] = "Buzz";
            else
                result[i - 1] = to_string(i);
        }
        return result;
    }
};

📸 Submission Screenshot:

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✅ Key Takeaways:

• Condition ordering matters: check for 3 & 5 first

• % operator helps easily detect multiples

• String conversion using to_string() for fallback case

• Indexing starts from i - 1 because result is 0-indexed

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📈 Daily Summary

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🏷️ Tags:
#leetcode #43DaysChallenge #fizzbuzz #cpp #strings #loops #conditionals #dsa