FindStartIndex(char[] a)
{
int start = 0;
int current = 1;
while(current < a.Length)
{
if(a[current] < a[start])
{
start = current;
++current;
}else if(a[current] > a[start])
{
++current;
}else //a[current] == a[start]
{
int lookforward = 0;
int startnext = start;
int currentnext = current;
while(startnext != currnet && a[startnext] == a[currentnext])
{
++lookforward;
startnext = (start + lookforward) % a.Length;
currentnext = (current + lookforward) % a.Length;
}//finish when compare to current head or there is different
if(startnext == current || a[startnext] < a[currentnext])
{
if(current > currentnext)
{
break;
}else
{
current = currentnext+1;
}
}
else //a[startnext] > a[currentnext]
{
start = current;
if(current < currentnext)
{
current = currentnext + 1;
}else
{
break;
}
}
}
}
return start;
}
On Fri, Jan 14, 2011 at 12:45 AM, radha krishnan <
radhakrishnance...@gmail.com> wrote:
> There s O(n) solution for this :)
>
> On Fri, Jan 14, 2011 at 2:13 PM, radha krishnan
> <radhakrishnance...@gmail.com> wrote:
> > append the string to original string and
> > index=answer of that spoj problem
> > now u can ouput the string from index to index+strlen(originalstring)-1
> >
> > On Fri, Jan 14, 2011 at 2:12 PM, radha krishnan
> > <radhakrishnance...@gmail.com> wrote:
> >> wow
> >> This s a spoj problem
> >> http://www.spoj.pl/problems/MINMOVE/
> >>
> >> On Fri, Jan 14, 2011 at 1:40 PM, snehal jain <learner....@gmail.com>
> wrote:
> >>> Write the code to find lexicographic minimum in a circular array, e.g.
> >>> for the array
> >>> BCABDADAB, the lexicographic mininum is ABBCABDAD.
> >>>
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