Finding all combinations of well-formed brackets

Question

This came up while talking to a friend and I thought I\'d ask here since it\'s an interesting problem and would like to see other people\'s solutions.

The task is to

Answer 1

Not the most elegant solution, but this was how I did it in C++ (Visual Studio 2008). Leveraging the STL set to eliminate duplicates, I just naively insert new () pairs into each string index in every string from the previous generation, then recurse.

#include "stdafx.h"
#include <iostream>
#include <string>
#include <set>

using namespace System;
using namespace std;

typedef set<string> StrSet;

void ExpandSet( StrSet &Results, int Curr, int Max )
{
    if (Curr < Max)
    {
        StrSet NewResults;

        for (StrSet::iterator it = Results.begin(); it != Results.end(); ++it)
        {
            for (unsigned int stri=0; stri < (*it).length(); stri++)
            {
                string NewStr( *it );
                NewResults.insert( NewStr.insert( stri, string("()") ) );
            }
        }
        ExpandSet( NewResults, Curr+1, Max );

        Results = NewResults;
    }
}    

int main(array<System::String ^> ^args)
{
    int ParenCount = 0;

    cout << "Enter the parens to balance:" << endl;
    cin  >> ParenCount;

    StrSet Results;
    Results.insert( string("()") );

    ExpandSet(Results, 1, ParenCount);

    cout << Results.size() << ": Total # of results for " << ParenCount << " parens:" << endl;

    for (StrSet::iterator it = Results.begin(); it != Results.end(); ++it)
    {
        cout << *it << endl;
    }


    return 0;
}

Answer 2

public static void printAllValidBracePermutations(int size) {
    printAllValidBracePermutations_internal("", 0, 2 * size);
}

private static void printAllValidBracePermutations_internal(String str, int bal, int len) {
    if (len == 0) System.out.println(str);
    else if (len > 0) {
        if (bal <= len / 2) printAllValidBracePermutations_internal(str + "{", bal + 1, len - 1);
        if (bal > 0) printAllValidBracePermutations_internal(str + "}", bal - 1, len - 1);
    }
}

Answer 3

def @memo brackets ( n )
    => [] if n == 0 else around( n ) ++ pre( n ) ++ post( n ) ++ [ "()" * n) ]

def @memo pre ( n )
    => map ( ( s ) => "()" ++ s, pre ( n - 1 ) ++ around ( n - 1 ) ) if n > 2 else []

def @memo post ( n )
    => map ( ( s ) => s ++ "()", post ( n - 1 ) ++ around ( n - 1 ) ) if n > 2 else []

def @memo around ( n )
    => map ( ( s ) => "(" ++ s ++ ")", brackets( n - 1 ) )

(kin, which is something like an actor model based linear python with traits. I haven't got round to implementing @memo but the above works without that optimisation)

Answer 4

Provider C# version based on recursive backtracking algorithm, hope it's helpful.

public List<String> generateParenthesis(int n) {
   List<String> result = new LinkedList<String>();
   Generate("", 0, 0, n, result);
   return result;
}

private void Generate(String s, int l, int r, int n, List<String> result){
   if(l == n && r == n){
       result.add(s);
       return;
   }

   if(l<n){
       Generate(s+"(", l+1, r, n, result);    
   }

   if(r < l)
       Generate(s+")", l , r+1, n, result);
 }}

Answer 5

Common Lisp:

This doesn't print them, but does produce a list of lists of all the possible structures. My method is a bit different from the others'. It restructures the solutions to brackets(n - 1) such that they become brackets(n). My solution isn't tail recursive, but it could be made so with a little work.

Code

(defun brackets (n)
  (if (= 1 n)
      '((()))
      (loop for el in (brackets (1- n))
            when (cdr el)
            collect (cons (list (car el)) (cdr el))
            collect (list el)
            collect (cons '() el))))

Answer 6

Here is a solution in C++. The main idea that I use is that I take the output from the previous i (where i is the number of bracket pairs), and feed that as input to the next i. Then, for each string in the input, we put a bracket pair at each location in the string. New strings are added to a set in order to eliminate duplicates.

#include <iostream>
#include <set>
using namespace std;
void brackets( int n );
void brackets_aux( int x, const set<string>& input_set, set<string>& output_set );

int main() {
    int n;
    cout << "Enter n: ";
    cin >> n;
    brackets(n);
    return 0;
}

void brackets( int n ) {
    set<string>* set1 = new set<string>;
    set<string>* set2;

    for( int i = 1; i <= n; i++ ) {
        set2 = new set<string>;
        brackets_aux( i, *set1, *set2 );
        delete set1;
        set1 = set2;
    }
}

void brackets_aux( int x, const set<string>& input_set, set<string>& output_set ) {
    // Build set of bracket strings to print
    if( x == 1 ) {
        output_set.insert( "()" );
    }
    else {
        // For each input string, generate the output strings when inserting a bracket pair
        for( set<string>::iterator s = input_set.begin(); s != input_set.end(); s++ ) {
            // For each location in the string, insert bracket pair before location if valid
            for( unsigned int i = 0; i < s->size(); i++ ) {
                string s2 = *s;
                s2.insert( i, "()" );
                output_set.insert( s2 );
            }
            output_set.insert( *s + "()" );
        }
    }

    // Print them
    for( set<string>::iterator i = output_set.begin(); i != output_set.end(); i++ ) {
        cout << *i << "  ";
    }
    cout << endl;
}