Are there any constraints in the problem, because it seems straight forward.
if number of elements are 2n indexed from 0 to 2n-1
for i=0 to n-1:
new_array[i*2]=old_array[i];
new_array[i*2+1]=old_array[i+n];
On Mon, Feb 28, 2011 at 7:41 PM, bittu <shashank7andr...@gmail.com> wrote:
> @jalaj U needs to clarify becoz what i can say that dat is overwritten
> in ur explanation so we loosing the original data where we are saving
> when we swapping the elements ur explanation seems to be right but
> little confusing
>
> @ujjwal i haven't tested ur code but i think its O(n^2) then why not
> try for this 20 Line Simple code of O(n^2)
>
> Finally we wants O(n) better solution fro this which exist for
> this ???
>
> #include<string.h>
>
> void swap(char *c,char *p)
> {
> char tmp;
> tmp=*c;
> *c=*p;
> *p=tmp;
>
> }
>
> int main (int argc, char const* argv[])
> {
> char str[] = "1234abcd";
> int i,j;
> int len = strlen(str)/2;
> //swap str[len-1] and str[len] and so on
> for ( i = 0; i < len-1; i += 1) {
> for ( j = len-1-i; j <= len+i; j += 2)
> {
> printf("i=%d j=%d c1=%c \t c2=%c
> \n",i,j,str[j],str[j+1]);
> swap(&str[j],&str[j+1]);
> }
> }
> printf("%s \n", str);
> return 0;
> }
>
> Time Complexcity O(n^2) O(n) Needed..????
>
>
> Thanks & Regards
> Shashank >> "The Best Way to escape From The Problem is Solve It"
>
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