LeetCode //C - 412. Fizz Buzz

412. Fizz Buzz

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.

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 < = 1 0 4 1 <= n <= 10^4 1<=n<=104

From: LeetCode

Link: 412. Fizz Buzz

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Solution:

Ideas:

1. Memory Allocation:

• The function returns an array of strings, so we first allocate memory for the array result that will hold the strings.
• Each string requires memory allocation separately. Since the maximum string length is 8 characters (for "FizzBuzz") plus 1 for the null terminator (\0), we allocate space for each string using malloc(9 * sizeof(char)).

2. FizzBuzz Logic:

• For each number from 1 to n, we check:
  • If the number is divisible by both 3 and 5, we assign "FizzBuzz".
  • If it's divisible by only 3, we assign "Fizz".
  • If it's divisible by only 5, we assign "Buzz".
  • Otherwise, we convert the number to a string using snprintf and store it in the array.

3. Returning the Result:

• The function returns the result array, and the size of the array (n) is returned through the returnSize pointer.
• The main function demonstrates how to call this function and free the allocated memory properly.

Code:

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
char** fizzBuzz(int n, int* returnSize) {
    *returnSize = n;
    char** result = (char**)malloc(n * sizeof(char*)); // Allocate memory for the result array

    for (int i = 1; i <= n; i++) {
        result[i - 1] = (char*)malloc(9 * sizeof(char)); // Allocate memory for each string (maximum length is 8 for "FizzBuzz" + 1 for '\0')

        if (i % 3 == 0 && i % 5 == 0) {
            strcpy(result[i - 1], "FizzBuzz");
        } else if (i % 3 == 0) {
            strcpy(result[i - 1], "Fizz");
        } else if (i % 5 == 0) {
            strcpy(result[i - 1], "Buzz");
        } else {
            snprintf(result[i - 1], 9, "%d", i); // Convert the integer to a string
        }
    }

    return result;
}