optimization question

[Brian Quinlan]

> >         def eqsub(s, i, j, t):
> >                 return (len(t) == j-i) and s[i:j] == t
> >
> > which avoids building the substrings unless necessary.
> >
> 
> Wouldn't the following avoid it altogether?
> 
>     return (len(t) == j-i) and (s.find(t,i) != -1)

That code won't quite work. I believe that this code will though:

def eqsub(s, i, j, t): 
	return (len(t) == j-i) and (s.find(t,i,j) != -1)

I don't know if the length check saves you anything.

Cheers,
Brian