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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