C Exercises: Combinations of parentheses from n pairs
Write a C programming to generate all combinations of well-formed parentheses from n given pairs of parentheses.
Example: Input n = 5 Output: (()())(()) (()())()() (())((())) (())(()()) (())(())() (())()(()) (())()()() ()(((()))) ()((()())) ()((())()) ()((()))() ()(()(())) ()(()()()) ()(()())() ()(())(()) ()(())()() ()()((())) ()()(()()) ()()(())() ()()()(()) ()()()()()
C Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static char** generate_Parenthesis(int n, int* returnSize)
{
int left_num, right_num, cap = 5000, ctr = 0;
char *stack = malloc(2 * n + 1);
char **parentheses = malloc(cap * sizeof(char *));
char *p = stack;
left_num = right_num = 0;
stack[2 * n] = '\0';
while (p != stack || ctr == 0) {
if (left_num == n && right_num == n) {
parentheses[ctr] = malloc(2 * n + 1);
strcpy(parentheses[ctr], stack);
ctr++;
while (--p != stack) {
if (*p == '(') {
if (--left_num > right_num) {
*p++ = ')';
right_num++;
break;
}
} else {
right_num--;
}
}
} else {
/* forward */
while (left_num < n) {
*p++ = '(';
left_num++;
}
while (right_num < n) {
*p++ = ')';
right_num++;
}
}
}
*returnSize = ctr;
return parentheses;
}
int main(void)
{
int i, ctr;
i = 5;
char ** lists = generate_Parenthesis(i, &ctr);
for (i = 0; i < ctr; i++) {
printf("%s\n", lists[i]);
}
return 0;
}
Sample Output:
((((())))) (((()()))) (((())())) (((()))()) (((())))() ((()(()))) ((()()())) ((()())()) ((()()))() ((())(())) ((())()()) ((())())() ((()))(()) ((()))()() (()((()))) (()(()())) (()(())()) (()(()))() (()()(())) (()()()()) (()()())() (()())(()) (()())()() (())((())) (())(()()) (())(())() (())()(()) (())()()() ()(((()))) ()((()())) ()((())()) ()((()))() ()(()(())) ()(()()()) ()(()())() ()(())(()) ()(())()() ()()((())) ()()(()()) ()()(())() ()()()(()) ()()()()()
Flowchart:
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