Generate Parentheses

#1

Test cases giving wrong answers, couldnt find the mistake

#2 
#include<iostream>
#include<algorithm>
#include<string>
#include<stdio.h> 
#include<stdlib.h>
#include<cstring>
#include<ctype.h>
#include<math.h>
#include<vector>
#include<time.h>
using namespace std;
typedef long long int ll;
bool printsol(char a[],int n,int i)
{
	//base
	if(i==2*n-1)
	{
		//print
		cout<<a<<endl;
		return false;
	}
	//no swap
	if(printsol(a,n,i+2))
	return true;
	//swap
	for(int k=i+1;k<2*n;k+=2)
	{	
		if(a[k]=='(')
		{
		swap(a[i],a[k]);
		if(printsol(a,n,i+2))
		return true;
		swap(a[i],a[k]);	
		}
		
	}
	return false;
}
int main() {
char a[23];
int n;
cin>>n;
for(int i=0;i<2*n;i+=2)
{
	a[i]='(';
	a[i+1]=')';
}
a[2*n]='\0';
bool p=printsol(a,n,1);
return 0;
}
#3

Hello @saarthakrox29,

  1. Share you code through an online IDE (Coding Blocks IDE).

  2. Following will help in correcting the code:
    Take two variables to count the occurrence of closing and opening bracket.
    Recursive APProach:
    2.1. base case i.e. printing the balanced string of parenthesis
    when close == n
    2.2. insert a closing ‘)’ parenthesis if there is an opening ‘(’ parenthesis already present in the output string or char array.
    when open > close
    2.3. insert a opening ‘(’ parenthesis if we can insert more brakets.
    when open < n

    the position of 2.2. and 2.3. will affect the order in which the balanced parenthesized will appear.
    Suggestion:
    Try to swap their positions to understand it better.

Hope, this would help.
Give a like if you are satisfied.

1 Like
#4

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