Fizz Buzz

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

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

                                            
class Solution(object):
    def fizzBuzz(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        result = []
        for i in range(1, n + 1):
            if i % 3 == 0 and i % 5 == 0:
                result.append("FizzBuzz")
            elif i % 3 == 0:
                result.append("Fizz")
            elif i % 5 == 0:
                result.append("Buzz")
            else:
                result.append(str(i))
        return result

                                            
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
 
char ** fizzBuzz(int n, int* returnSize){
    *returnSize = n;
    char **answer = (char **)malloc(n * sizeof(char *));
    for(int i = 1; i <= n; i++){
        if(i % 3 == 0 && i % 5 == 0){
            answer[i-1] = "FizzBuzz";
        } else if(i % 3 == 0){
            answer[i-1] = "Fizz";
        } else if(i % 5 == 0){
            answer[i-1] = "Buzz";
        } else {
            answer[i-1] = (char *)malloc(10 * sizeof(char));
            sprintf(answer[i-1], "%d", i);
        }
    }
    return answer;
}

                                            
public class Solution {
    public IList<string> FizzBuzz(int n) {
        IList<string> answer = new List<string>();
        
        for(int i=1; i<=n; i++){
            if(i % 3 == 0 && i % 5 == 0){
                answer.Add("FizzBuzz");
            }
            else if(i % 3 == 0){
                answer.Add("Fizz");
            }
            else if(i % 5 == 0){
                answer.Add("Buzz");
            }
            else{
                answer.Add(i.ToString());
            }
        }
        
        return answer;
    }
}

                                            
/**
 * @param {number} n
 * @return {string[]}
 */
var fizzBuzz = function(n) {
    const result = [];
    for (let i = 1; i <= n; i++) {
        if (i % 3 === 0 && i % 5 === 0) {
            result.push("FizzBuzz");
        } else if (i % 3 === 0) {
            result.push("Fizz");
        } else if (i % 5 === 0) {
            result.push("Buzz");
        } else {
            result.push(i.toString());
        }
    }
    return result;
};

                                            
function fizzBuzz(n: number): string[] {
    const result: string[] = [];
    
    for (let i = 1; i <= n; i++) {
        if (i % 3 === 0 && i % 5 === 0) {
            result.push("FizzBuzz");
        } else if (i % 3 === 0) {
            result.push("Fizz");
        } else if (i % 5 === 0) {
            result.push("Buzz");
        } else {
            result.push(String(i));
        }
    }
    
    return result;
};

                                            
class Solution {
    /**
     * @param Integer $n
     * @return String[]
     */
    function fizzBuzz($n) {
        $res = [];
        for ($i = 1; $i <= $n; $i++) {
            if ($i % 3 == 0 && $i % 5 == 0) {
                $res[] = "FizzBuzz";
            } elseif ($i % 3 == 0) {
                $res[] = "Fizz";
            } elseif ($i % 5 == 0) {
                $res[] = "Buzz";
            } else {
                $res[] = (string) $i;
            }
        }
        return $res;
    }
}

                                            
class Solution {
    func fizzBuzz(_ n: Int) -> [String] {
        var answer = [String]()
        
        for i in 1...n {
            if i % 3 == 0 && i % 5 == 0 {
                answer.append("FizzBuzz")
            } else if i % 3 == 0 {
                answer.append("Fizz")
            } else if i % 5 == 0 {
                answer.append("Buzz")
            } else {
                answer.append(String(i))
            }
        }
        
        return answer
    }
}

To solve this coding challenge, we need to create an algorithm that processes a given integer

n

and generates a list of strings based on specific divisibility rules. This list will contain the numbers from 1 to

n

, but certain numbers will be replaced by the words "Fizz," "Buzz," or "FizzBuzz" according to the rules described in the problem statement.

Explanation

The problem is to implement the FizzBuzz game, a well-known programming exercise. The rules are straightforward:

For numbers that are divisible by both 3 and 5, the number should be replaced by "FizzBuzz".
For numbers that are divisible by 3 only, the number should be replaced by "Fizz".
For numbers that are divisible by 5 only, the number should be replaced by "Buzz".
If the number is not divisible by 3 or 5, it should remain as is but converted to a string.

Given these rules, our approach will be:

Initialize an empty list to store the result.
Iterate through a range of numbers from 1 to
n

.
For each number, check the divisibility:

If the number is divisible by both 3 and 5, append "FizzBuzz" to the list.
If the number is divisible by only 3, append "Fizz" to the list.
If the number is divisible by only 5, append "Buzz" to the list.
If none of the conditions above are met, append the number itself as a string.

Return the list at the end.

Step-by-Step Explanation

Now let's break this down further into detailed steps:

Step-by-Step Pseudocode Explanation

Initialization:

Create an empty list called
result

to store the final output.

Loop through Numbers:

Iterate over a range from 1 through
n

(inclusive).

Check Divisibility:

For each number in the range:

If the number is divisible by both 3 and 5:

Append "FizzBuzz" to the
result

list.

Else, if the number is divisible by 3:

Append "Fizz" to the
result

list.

Else, if the number is divisible by 5:

Append "Buzz" to the
result

list.

Else:

Append the number itself (converted to string) to the
result

list.

Output:

Return the
result

list.

Detailed Steps in Pseudocode

                                            
# Initialize the result list to store the output
result = []

# Loop through numbers from 1 to n (inclusive)
for i from 1 to n:
    
    # Check divisibility for both 3 and 5
    if i % 3 == 0 and i % 5 == 0:
        # Append "FizzBuzz" if divisible by both 3 and 5
        result.append("FizzBuzz")
    
    # Check divisibility for 3
    else if i % 3 == 0:
        # Append "Fizz" if divisible by 3
        result.append("Fizz")

    # Check divisibility for 5
    else if i % 5 == 0:
        # Append "Buzz" if divisible by 5
        result.append("Buzz")
    
    # If none of the above, append the number itself
    else:
        # Convert the number to string and append
        result.append(str(i))

# Return the resulting list containing FizzBuzz sequence
return result

Thus, following these steps in the pseudocode, we can construct our solution to the FizzBuzz problem. Each step carefully accounts for the rules provided and ensures all edge cases are handled seamlessly by checking divisibility conditions in sequence. Implementing this algorithm ensures that we adhere to the problem constraints and achieve the desired result efficiently.