[Easy] Bracket Simplification

Welcome to the first programming challenge! Three of these will be posted a week and you can complete it in any language you want.

You get a point for completing an easy challenge, 2 for a medium, and 3 for a hard. For each challenge if you solve it in the least amount of characters you get a bonus point, and if your code runs the fastest when I check it you also get a bonus point. (ties mean everyone who tied gets the bonus point although exact duplicate answers wont count)

Ill be posting a leaderboard that will show the people who have the most points every month

Submissions will be open for a week

As a new hire of bracket inc., you have been tasked with getting rid of excess brackets lying around the facility. You must simplify a series of brackets so that only brackets that dont have a match remain (a match is an opening and closing bracket of the same type beside each other). The final result should have no matches

As an example for the input [(({})({)(()}] the expected output would be [(({)(}]

These are the valid types of brackets: (){}[]

Your system will be tested against 10 different unknown test cases before it is unleashed on the facility. In order to complete this task you must pass all of the test cases.

Any programming language may be used and to submit an answer reply on this post with the code and the language you coded it in

Edit: Clarification, you must take input in from the user using the program instead of them being hardcoded. (makes it easier to test)

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shape-warrior-t, 10 months ago
Here's an O(n) solution using a stack instead of repeated search & replace:
closing_to_opening = {')': '(', ']': '[', '}': '{'}
brackets = input()
acc = []
for bracket in brackets:
    if bracket in closing_to_opening:
        if acc and acc[-1] == closing_to_opening[bracket]:
            acc.pop()
        else:
            acc.append(bracket)
    else:
        acc.append(bracket)
print(''.join(acc))
Haven't thoroughly thought the problem through (so I'm not 100% confident in the correctness of the solution), but the general intuition here is that pairs of brackets can only match up if they only have other matching pairs of brackets between them. You can deal with matching pairs of brackets on the fly simply by removing them, so there's actually no need for backtracking.

Golfed, just for fun:
a=[]
[a.pop()if a and a[-1]==dict(zip(')]}','([{')).get(b)else a.append(b)for b in input()]
print(''.join(a))
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brie, 10 months ago

This gave me the idea to do the same in C, but use the argument string itself as the stack to avoid any allocations. It could probably be further optimized.


<span style="color:#323232;">#include <stdio.h>
</span><span style="color:#323232;">
</span><span style="color:#323232;">int main(int argc, char **argv)
</span><span style="color:#323232;">{
</span><span style="color:#323232;">	char map[256] = { 0 };
</span><span style="color:#323232;">	map[')'] = '(';
</span><span style="color:#323232;">	map['}'] = '{';
</span><span style="color:#323232;">	map[']'] = '[';
</span><span style="color:#323232;">
</span><span style="color:#323232;">	while (--argc) {
</span><span style="color:#323232;">		char *top, *p, c;
</span><span style="color:#323232;">		top = p = *++argv;
</span><span style="color:#323232;">
</span><span style="color:#323232;">		while ((c = *p++)) {
</span><span style="color:#323232;">			if (top != *argv && map[(size_t)c] == top[-1]) {
</span><span style="color:#323232;">				top--;
</span><span style="color:#323232;">			} else {
</span><span style="color:#323232;">				*top++ = c;
</span><span style="color:#323232;">			}
</span><span style="color:#323232;">		}
</span><span style="color:#323232;">
</span><span style="color:#323232;">		*top = 0;
</span><span style="color:#323232;">
</span><span style="color:#323232;">		puts(*argv);
</span><span style="color:#323232;">	}
</span><span style="color:#323232;">}
</span>

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