One solution to this problem is that :
char s[]="abracadbra";
char p[]="bca";
char temp[];
1. First Remove element b of array p from array s .
Compare b of array p with with characters of array s .
if b is not matched with characters of array s
insert those characters of s into new temp array .
continue this till s reaches '\0';
now again reassign the reduced characters of temp array back into
array s;
continuously repeat the step 1 for another characters in p till p become or
reaches '\0' ;
finally you have an array which doesn't have characters which are there in p
array;
I think it only need an extra temporary array , but at each interation
array size is getting reduced & as well no. of time to compare elements is
also get reduced .
If any one have questions please ask;
Thank you ,
---
Regards
Peeyush Bishnoi
On 8/7/07, Arulanandan P <[EMAIL PROTECTED]> wrote:
>
> You have to write a function whose prototype is given bellow. this
> function will accept two char * named subject and pattern. for example
> subject="abracadbra"
> and pattern="bca".now it should check occurrences of all chars of string
> pattern in subject . If any match occurs then it will remove that char from
> subject . so finally , as in our example
> at end subject ="rdr"
>
> void fun(char *subject,char *pattern)
> {
> // write your code here
> }
> >
>
--
----
Peeyush Bishnoi
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