how abt dis guys ?? #include <stdio.h> #include <string.h> #define MAX 100
int main() { int n; int i; int j; int it; char input[MAX]; char tmp; scanf("%s",input); n = strlen(input); i = j = n/2; for(it=1; it<n-2; it++) { if(it%2 == 1) { tmp = input[j]; input[j] = input[it]; input[it] = tmp; j++; } else if( it < i ) { tmp = input[i]; input[i] = input[it]; input[it] = tmp; } else { tmp = input[it]; input[it] = input[it+1]; input[it+1] = tmp; } } printf("\n%s\n",input); return 0; } On Mon, Feb 28, 2011 at 8:29 PM, Arpit Sood <soodfi...@gmail.com> wrote: > 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" >> >> -- >> 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. >> >> > > -- > 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. > -- 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.