@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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