Fizz Buzz(LeetCode)

Problem Statement

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.

Examples

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

The FizzBuzz problem is a classic programming task often used to teach and test fundamental coding skills. The problem statement is simple: for a given integer n, return a list of strings where each string corresponds to the numbers from 1 to n with the following rules:

• For multiples of 3, add "Fizz" instead of the number.
• For multiples of 5, add "Buzz" instead of the number.
• For multiples of both 3 and 5, add "FizzBuzz" instead of the number. Here, we present three different approaches to solve the FizzBuzz problem in Java.

Approach 1: Simple if-else

Algorithm

1. Create an empty list to store the results.
2. Loop from 1 to n.
3. For each number, check if it is divisible by 15, 3, or 5 using the modulus operator.
4. Add the appropriate string ("FizzBuzz", "Fizz", "Buzz", or the number itself) to the list.
5. Return the list.

Implementation

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            if (i % 15 == 0) {
                ans.add("FizzBuzz");
            } else if (i % 3 == 0) {
                ans.add("Fizz");
            } else if (i % 5 == 0) {
                ans.add("Buzz");
            } else {
                ans.add(String.valueOf(i));
            }
        }
        return ans;                 
    }
}

Approach 2: No Modulus Operator

Algorithm

1. Create an empty list to store the results.
2. Use two counters, fizz and buzz, initialized to 0.
3. Loop from 1 to n.
4. Increment fizz and buzz for each number.
5. Check the values of fizz and buzz to decide whether to add "Fizz", "Buzz", "FizzBuzz", or the number itself to the list.
6. Reset fizz and/or buzz when they reach their respective thresholds (3 and 5).
7. Return the list.

Implementation

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> res = new ArrayList<>();
        int i = 1, fizz = 0, buzz = 0;
        while (i <= n) {
            fizz++; buzz++;
            if (fizz == 3 && buzz == 5) {
                res.add("FizzBuzz");
                fizz = buzz = 0;
            } else if (fizz == 3) {
                res.add("Fizz");
                fizz = 0;
            } else if (buzz == 5) {
                res.add("Buzz");
                buzz = 0;
            } else {
                res.add(String.valueOf(i));
            }
            i++;
        }
        return res;
    }
}

Approach 3: Ternary Operator

Algorithm

1. Create an empty list to store the results.
2. Loop from 1 to n.
3. For each number, use the ternary operator to check if it is divisible by 15, 5, or 3, and add the appropriate string to the list.
4. Return the list.

Implementation

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            ans.add(
                i % 15 == 0 ? "FizzBuzz" :
                i % 5 == 0  ? "Buzz" :
                i % 3 == 0  ? "Fizz" :
                String.valueOf(i)
            );
        }
        return ans;
    }
}

Complexity Analysis

• Time complexity: $O(N)$
• Space complexity: $O(N)$

Conclusion

The choice of approach can depend on coding style and specific constraints or preferences. The first approach is straightforward and easy to understand. The second approach is a creative way to avoid the modulus operator, while the third approach uses the ternary operator for a concise solution.