bool interleave(string s1,string s3)
{
char *str1=(char*)s1.c_str();
char *str3=(char*)s3.c_str();
int pos=-1;
for(int i=0;i<strlen(str1);i++)
{
if(pos<findPosition(str1[i],str3))
{
pos=findPosition(str1[i],str3);
}
else {flag=1; break;}
}
if(flag==1)
print NOT........
}
Do the same for string2 also ...
This works only for non duplicated strings ....
On Thu, Sep 1, 2011 at 5:19 AM, vikas <[email protected]> wrote:
> @ above, a third string is used, s3 which is strlen(s2)+ strlen(s1)
> and thus in O(n) space
>
> I guess qs calls out for O(1) space.
>
> besides , if we have O(n) space , this question simply reduces to
> finding the number of permutation of string s1+s2
>
> I doubt we can do it in O(1) space, any idea guys ???
>
>
> On Aug 31, 7:00 pm, Don <[email protected]> wrote:
> > // Returns true if string s3 is s1 interleaved with s2.
> > // The function is iterative when possible, but uses a recursive
> > // call when both s1 and s2 match the next character in s3.
> > // Note that this function is not intended to be called directly. It
> > is called by Interleaved().
> > bool interleaved2(char *s1, char *s2, char *s3)
> > {
> > while(1)
> > {
> > // End case is when all three strings are empty
> > if (!s1[0] && !s2[0] && !s3[0]) return true;
> > if (s1[0] == s3[0])
> > {
> > if (s2[0] == s3[0])
> > {
> > // Tries both using s1 and s2 next. The recursive call uses
> > s1,
> > // and the postincrement of s2 uses s2 iteratively.
> > if (interleaved2(s1+1, s2++,s3+1)) return true;
> > }
> > // s1 is the only match
> > else ++s1;
> > }
> > else
> > {
> > // s2 is the only match
> > if (s2[0] == s3[0]) ++s2;
> >
> > // Neither s1 nor s2 match the next character in s3 so the
> > strings are not interleaved
> > else return false;
> > }
> >
> > // Move on to the next character in s3
> > ++s3;
> > }
> >
> > }
> >
> > bool Interleaved(char *s1, char *s2, char *s3)
> > {
> > // Frequency counts
> > int count1[256] = {0};
> > int count2[256] = {0};
> > int i,j;
> >
> > // Count the number of occurances of each character in s1 and s2
> > for(i = 0; s1[i]; ++i)
> > ++count1[s1[i]];
> > for(j = 0; s2[j]; ++j)
> > ++count1[s2[j]];
> > j += i;
> >
> > // Next count the number of occurances of each character in s3
> > for(i = 0; s3[i]; ++i)
> > ++count2[s3[i]];
> >
> > // If the total number of characters in s3 is not s1+s2, interleaved
> > is false
> > if (i != j) return false;
> >
> > // If s3 is s1 interleaved with s2, these counts must be equal
> > for(i = 1; i < 128; ++i)
> > if (count1[i] != count2[i])
> > return false;
> >
> > // Call the function to do the real work.
> > return interleaved2(s1,s2,s3);
> >
> > }
> >
> > On Aug 31, 8:43 am, Navneet <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Suppose the two strings are ab and cd.
> >
> > > The possible strings formed by interleaving these two are
> > > abcd, acbd, acdb , cabd etc..
> >
> > > On Aug 31, 5:23 pm, sukran dhawan <[email protected]> wrote:
> >
> > > > what do u mean by interleaving ?
> >
> > > > On Wed, Aug 31, 2011 at 5:01 PM, Navneet Gupta <
> [email protected]>wrote:
> >
> > > > > The important thing to notice here is that relative order of
> > > > > characters is important and hence you should not look for just
> count
> > > > > char based approaches.
> >
> > > > > On Wed, Aug 31, 2011 at 11:20 AM, Navneet Gupta <
> [email protected]>
> > > > > wrote:
> > > > > > Given two strings S1 and S2, Find whether another string S3 can
> formed
> > > > > > by interleaving S1 and S2. Only constant space.
> >
> > > > > > --
> > > > > > Regards,
> > > > > > Navneet
> >
> > > > > --
>
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