Even my recursive solution will not work. :). Nice catch.!!.
int interleaved(char *s1, char *s2, char *s3) {
if (s1 == null && s2== null && s3==null) return 1;
if (s3==null) return 0;
if (s1 != null && *s1 == *s3 && s2 != null && *s2 == *s3) return
interleaved(s1+1,s2,s3+1) || return interleaved(s1, s2+1,s3+1);
if (s1 != null && *s1 == *s3) return interleaved(s1+1,s2,s3+1);
else if (s2 != null && *s2 == *s3) return interleaved(s1, s2+1,s3+1);
return 0;
}
int interleaved(char *s1, char *s2, char *s3) {
if (s1 == null && s2== null && s3==null) return 1;
if (s3==null) return 0;
while (*s3) {
// Added code here. So now its no more a iterative solution.
if (s1 != null && *s1 == *s3 && s2 != null && *s2 == *s3) return
interleaved(s1+1,s2,s3+1) || return interleaved(s1, s2+1,s3+1);
if (s1 != null && *s1 == *s3) {
s1++;
}
else if (s2 != null && *s2 == *s3) {
s2++;
} else {
return 0;
}
s3++;
}
return (s1 == null) && (s2 == null) && (s3 == null);
}
On Sat, May 21, 2011 at 1:05 AM, Don <dondod...@gmail.com> wrote:
> Your iterative solution does not work in cases where both s1 and s2
> have the next character in s3, but only choosing the s2 character next
> will result in correct interleaving.
>
> s1 = "ab"
> s2 = "axb"
> s3 = "axabb"
>
> Your iterative solution will say that these are not interleaved, but
> they really are.
>
> Don
>
> On May 20, 2:21 pm, immanuel kingston <kingston.imman...@gmail.com>
> wrote:
> > A Recursive solution:
> >
> > int interleaved(char *s1, char *s2, char *s3) {
> > if (s1 == null && s2== null && s3==null) return 1;
> > if (s3==null) return 0;
> > if (s1 != null && *s1 == *s3) return interleaved(s1+1,s2,s3+1);
> > else if (s2 != null && *s2 == *s3) return interleaved(s1,
> s2+1,s3+1);
> > return 0;
> >
> > }
> >
> > Iterative solution:
> > int interleaved(char *s1, char *s2, char *s3) {
> > if (s1 == null && s2== null && s3==null) return 1;
> > if (s3==null) return 0;
> > while (*s3) {
> > if (s1 != null && *s1 == *s3) {
> > s1++;
> > }
> > else if (s2 != null && *s2 == *s3) {
> > s2++;
> > } else {
> > return 0;
> > }
> > s3++;
> > }
> > return (s1 == null) && (s2 == null) && (s3 == null);
> >
> > }
> >
> > Thanks,
> > Immanuel
> >
> > On Fri, May 20, 2011 at 8:42 PM, Don <dondod...@gmail.com> wrote:
> > > This is the same algorithm as my previous solution. Both produce the
> > > correct result, but this one does not have tail recursion, so it will
> > > run faster.
> >
> > > bool interleaved2(char *s1, char *s2, char *s3)
> > > {
> > > while(1)
> > > {
> > > if (!s1[0] && !s2[0] && !s3[0]) return true;
> > > if (s1[0] == s3[0])
> > > {
> > > if (s2[0] == s3[0])
> > > {
> > > if (interleaved2(s1+1, s2++,s3+1)) return true;
> > > }
> > > else ++s1;
> > > }
> > > else
> > > {
> > > if (s2[0] == s3[0]) ++s2;
> > > else return false;
> > > }
> > > ++s3;
> > > }
> > > }
> >
> > > On May 19, 1:33 pm, Piyush Sinha <ecstasy.piy...@gmail.com> wrote:
> > > > Design an algorithm to find whether a given string is formed by the
> > > > interleaving of two given strings or not. e.g. s1= aabccabc s2=
> dbbabc
> > > s3=
> > > > aabdbbccababcc Given s1,s2,s3 design an efficient algorithm to find
> > > whether
> > > > s3 is formed from the interleaving of s1 and s2
> >
> > > > --
> > > > *Piyush Sinha*
> > > > *IIIT, Allahabad*
> > > > *+91-8792136657*
> > > > *+91-7483122727*
> > > > *https://www.facebook.com/profile.php?id=100000655377926*
> >
> > > --
> > > You received this message because you are subscribed to the Google
> Groups
> > > "Algorithm Geeks" group.
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> >
> >
>
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