@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" -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.