LeetCode //C - 22. Generate Parentheses

22. Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
 

Example 1:

Input: n = 3
Output: [“((()))”,“(()())”,“(())()”,“()(())”,“()()()”]

Example 2:

Input: n = 1
Output: [“()”]

Constraints:

• 1 <= n <= 8

From: LeetCode
Link: 22. Generate Parentheses

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Solution:

Ideas：

The recursive approach is employed to generate all possible combinations. At each recursive call, the code decides whether to add an opening parenthesis ( or a closing parenthesis ) based on certain conditions.

Key Observations:
You can only add an opening parenthesis if the number used so far is less than n.
You can only add a closing parenthesis if the number of opening parentheses used so far is greater than the number of closing parentheses. This ensures that we never have a situation like ()).

Code:

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
void generate(int n, int open, int close, char* current, int index, char*** result, int* returnSize) {
    if (index == 2 * n) {
        current[index] = '\0';
        (*result)[*returnSize] = strdup(current);
        (*returnSize)++;
        return;
    }
    
    if (open < n) {
        current[index] = '(';
        generate(n, open + 1, close, current, index + 1, result, returnSize);
    }
    
    if (close < open) {
        current[index] = ')';
        generate(n, open, close + 1, current, index + 1, result, returnSize);
    }
}

int catalan(int n) {
    int result = 1;
    for (int i = 0; i < n; ++i) {
        result *= (2 * n - i);
        result /= (i + 1);
    }
    return result / (n + 1);
}

char** generateParenthesis(int n, int* returnSize) {
    *returnSize = 0;
    int maxCombination = catalan(n);
    char** result = (char**) malloc(maxCombination * sizeof(char*));
    char* current = (char*) malloc(2 * n + 1);
    
    generate(n, 0, 0, current, 0, &result, returnSize);
    
    free(current);
    return result;
}
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