On Fri, Sep 17, 2010 at 3:36 PM, Krunal Modi <krunalam...@gmail.com> wrote:
> Your solutions are pretty impressive.
> Which place(country) are you from ?
> where are you studying (or done :) ) ?
>
> Keep it up...
> Good Wishes..
> --Krunal
>
> On Sep 14, 9:29 pm, Gene <gene.ress...@gmail.com> wrote:
> > You can approach this the same way you'd do it by hand. Build up the
> > string of brackets left to right. For each position, you have a
> > decision of either ( or ) bracket except for two constraints:
> > (1) if you've already decided to use n left brackets, then you can't
> > use a another left bracket and
> > (2) if you've already used as many right as left brackets, then you
> > can't use another right one.
> >
> > This suggests the following alorithm. Showing what happens on the
> > stack is a silly activity.
> >
> > #include <stdio.h>
> >
> > // Buffer for strings of ().
> > char buf[1000];
> >
> > // Continue the printing of bracket strings.
> > // need is the number of ('s still needed in our string.
> > // open is tne number of ('s already used _without_ a matching ).
> > // tail is the buffer location to place the next ) or (.
> > void cont(int need, int open, int tail)
> > {
> > // If nothing needed or open, we're done. Print.
> > if (need == 0 && open == 0) {
> > printf("%s\n", buf);
> > return;
> > }
> >
> > // If still a need for (, add a ( and continue.
> > if (need > 0) {
> > buf[tail] = '(';
> > cont(need - 1, open + 1, tail + 1);
> > }
> >
> > // If still an open (, add a ) and continue.
> > if (open > 0) {
> > buf[tail] = ')';
> > cont(need, open - 1, tail + 1);
> > }
> >
> > }
> >
> > void Brackets(int n)
> > {
> > cont(n, 0, 0);
> >
> > }
> >
> > int main(void)
> > {
> > Brackets(3);
> > return 0;
> >
> > }
> >
> > On Sep 14, 10:57 am, bittu <shashank7andr...@gmail.com> wrote:
> >
> >
> >
> > > Write a function Brackets(int n) that prints all combinations of well-
> > > formed brackets. For Brackets(3) the output would be ((())) (()()) (())
> > > () ()(()) ()()()
> >
> > > with explaination dat is at every call what is contant of stack during
> > > pushing and popping
>
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> #include<iostream.h>
#include<conio.h>
#include<math.h>
void addb(int n,int cnt,int i,char *a)
{
if(n==0) {
{
while(cnt!=0){
a[i++]=')';
cnt--; }
//if(cnt==0)
{
// printf("%s\n",a);
cout<<a<<"\n";
}
return;
}
}
else
{
a[i]='(';
addb(n-1,cnt+1,i+1,a);
if(cnt!=0)
{
a[i]=')';
addb(n,cnt-1,i+1,a);
}
}
}
void main()
{
clrscr();
char *a;
int n;
cout<<"n = ";
cin>>n;
a=new char[n*2+1];
a[n*2]='\0';
addb(n,0,0,a);
getch();
}
--
Pratik Kathalkar
CoEP
BTech IT
8149198343
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