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[Med] Bracket Match Generation (programming.dev)

submitted 1 year ago* (last edited 1 year ago) by Ategon@programming.dev to c/challenges@programming.dev

17 comments fedilink hide all child comments

Bracket Inc. wants to ship out new products using their excess brackets. They have tasked you with generating every possible assortment of brackets for some n brackets where the brackets will match

A bracket match is an opening and closing version of the same kind of bracket beside each other ()
If a bracket matches then outer brackets can also match (())
n will be an even number
The valid brackets are ()[]{}

For example for n = 4 the options are

()()
(())
[][]
[[]]
{}{}
{{}}
[]()
()[]
(){}
{}()
[]{}
{}[]
({})
{()}
([])
[()]
{[]}
[{}]

You must accept n as a command line argument (entered when your app is ran) and print out all of the matches, one per line

(It will be called like node main.js 4 or however else to run apps in your language)

You can use the solution tester in this post to test you followed the correct format https://programming.dev/post/1805174

Any programming language may be used. 2 points will be given if you pass all the test cases with 1 bonus point going to whoevers performs the quickest and 1 for whoever can get the least amount of characters

To submit put the code and the language you used below

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[–] brie 1 points 1 year ago* (last edited 1 year ago)

Figured out my old solution didn't actually work past 4, so here's a new solution in C. It matches @SleveMcDichael@programming.dev's solution up to n=10.

Some stats using wc -l:

n=0 has 1 combinations
n=2 has 3 combinations
n=4 has 18 combinations
n=6 has 135 combinations
n=8 has 1134 combinations
n=10 has 10206 combinations
n=12 has 96228 combinations
n=14 has 938223 combinations
n=16 has 9382230 combinations
n=18 has 95698746 combinations
n=20 has 991787004 combinations

#include <stdio.h>
#include <stdlib.h>

char const brackets[6] = "({[)}]";

int update(int *buf, int len)
{
	int width = *buf >> 2;
	if (width > len - 2) {
		return -1;
	}
	if ((++*buf & 3) < 3) {
		return 0;
	}
	*buf &= ~3;

	if (update(buf + 1, width)) {
		buf[1] = 0;
		if (update(buf + width + 2, len - width - 2)) {
			*buf = (width += 2) << 2;
			if (width > len - 2) {
				return -1;
			}
			buf[width + 2] = 0;
		}
	}
	return 0;
}

void display(int *buf, int len)
{
	int width = *buf >> 2;
	char const *bracket = brackets + (*buf & 3);
	if (len <= 0) {
		return;
	};

	putchar(bracket[0]);
	display(buf + 1, width);
	putchar(bracket[3]);
	display(buf + width + 2, len - width - 2);
}

int main(int argc, char **argv)
{
	int n;
	int *buf;
	if (argc < 2) {
		fputs("Bad invocation", stderr);
		exit(EXIT_FAILURE);
	}

	sscanf(argv[1], "%d", &n);

	buf = calloc(n + 2, sizeof(*buf));

	display(buf, n);
	putchar('\n');
	while (!update(buf, n)) {
		display(buf, n);
		putchar('\n');
	}

	free(buf);
	return 0;
}

v1 slightly incorrect solution

#include <stdio.h>
#include <stdlib.h>

char const brackets[6] = "({[)}]";

int update(int *buf, int len)
{
	int width = *buf >> 2;
	if (width > len - 2) {
		return -1;
	}
	if ((++*buf & 3) < 3) {
		return 0;
	}
	*buf &= ~3;

	if (update(buf + 1, width)) {
		buf[1] = 0;
		if (update(buf + width + 2, len - width - 2)) {
			*buf = (width += 2) << 2;
			if (width > len - 2) {
				return -1;
			}
			buf[width + 2] = 0;
		}
	}
	return 0;
}

void display(int *buf, int len)
{
	int width = *buf >> 2;
	char const *bracket = brackets + (*buf & 3);
	if (len <= 0) {
		return;
	};

	putchar(bracket[0]);
	display(buf + 1, width);
	putchar(bracket[3]);
	display(buf + width + 2, len - width - 2);
}

int main(int argc, char **argv)
{
	int n;
	int *buf;

	sscanf(argv[1], "%d", &n);
	if (n == 0) {
		return 0;
	}

	buf = calloc(n + 20, sizeof(*buf));

	display(buf, n);
	putchar('\n');
	while (!update(buf, n)) {
		display(buf, n);
		putchar('\n');
	}

	free(buf);
	return 0;
}