it can be solved in O(n).
let query array be b[k] and array of int is a[n],
int j=i=0, s=-1;
for(i=0;i<n;i++)
{
if(a[i]==b[j])
{
j++;
if(j==1)
s=i;
}
else
{
j=0;
s=-1;
}
if(j==k)
break;
}
if( i<=n && j==k)
printf("Found starting index=%d",s);
else
printf("NOt found");
return 0;
}
On Oct 7, 1:20 pm, RAHUL KUJUR <kujurismonu2...@gmail.com> wrote:
> @prodigy: how is it coming O(nlogk) can u explain???
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