Objective

The objective of this program is to generate all valid combinations of n pairs of parentheses.
Valid combinations are those where each opening parenthesis ‘(‘ has a corresponding closing parenthesis ‘)’,
and the order is such that no closing parenthesis appears before its matching opening parenthesis.

Code

#include 
#include 
#include 

void generateParenthesisUtil(char *current, int pos, int open, int close, int n) {
    if (close == n) {
        printf("%s\n", current);
        return;
    }
    if (open < n) {
        current[pos] = '(';
        generateParenthesisUtil(current, pos + 1, open + 1, close, n);
    }
    if (close < open) {
        current[pos] = ')';
        generateParenthesisUtil(current, pos + 1, open, close + 1, n);
    }
}

void generateParenthesis(int n) {
    char *current = (char *)malloc(2 * n + 1);
    current[2 * n] = '\0';
    generateParenthesisUtil(current, 0, 0, 0, n);
    free(current);
}

int main() {
    int n;
    printf("Enter the number of pairs of parentheses: ");
    scanf("%d", &n);
    generateParenthesis(n);
    return 0;
}

Program Structure

The program consists of the following main components:

• Function generateParenthesisUtil: This is a recursive function that builds valid combinations of parentheses. It takes the current combination string, its position, counts of open and close parentheses, and the total pairs n as parameters.
• Function generateParenthesis: This function initializes the current combination string and calls the utility function.
• Main Function: This is the entry point of the program where user input is taken for the number of pairs, and the generation function is invoked.

How to Run the Program

1. Copy the code into a C compiler or IDE (e.g., GCC, Code::Blocks, Dev-C++).
2. Compile the program to create the executable.
3. Run the executable.
4. When prompted, enter the number of pairs of parentheses you wish to generate.
5. The program will output all valid combinations of the specified number of parentheses.

Example

If you enter 3, the output will be:

((()))
(()())
(())()
()(())
()()()