@Dhilip
Is it tested ? I doubt your code won't work ?
@Rohit
Can we anyways modify Morris Inorder Traversal process? We can have two
pointers slow(increments once) and fast(increments twice), so that if fast
reaches end or fast->next is end, we can have the median @ slow ?
Correct me If I am wrong
Sharing link for Morris Inorder
http://www.scss.tcd.ie/disciplines/software_systems/fmg/fmg_web/IFMSIG/winter2000/HughGibbonsSlides.pdf
Courtesy Rohit
On Mon, May 17, 2010 at 3:42 PM, kaushik sur wrote:
> @Manish
>
> Does not a recursive solution [inorder traversal] takes an impli
@Manish
Does not a recursive solution [inorder traversal] takes an implicit stack
space ?
Please correct me if I am wrong ?
@Rahul
Can you please send us the *morris inorder pdf* link that u have shared once
?
Best Regards
Kaushik
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Hi Friends
Hash Map takes 2byte [in Java] for holding a character
So in Amazon -
It takes A - 1
M - 1
Z - 1
O - 1
N - 1
But it's time effective!
Yes it takes additional space for intergers, for each key 4 byte for an
integer!!! :-(
***
public void checkTheFrequency() {
Thanks Rohit!
On Sat, May 15, 2010 at 7:13 PM, Rohit Saraf wrote:
> there is something called morris inorder traversal.
> credits to donald knuth
>
>
> On 5/15/10, kaushik sur wrote:
> > Hi Friends
> >
> > I have encountered the question in sites - "G
r and keeping a
non-static count variable.
I welcome any better solution, time and space efficient.
Best Regards
Kaushik Sur
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Hi Friend
Using HashMap in Java
***
/*
*
input a character array from the user and output in the following way.
example string is amazon.. then output should be a2m1z1o1n1
*/
package questionaire;
import java.util.Collection;
import java.util.HashMap;
impo
gs) {
Amazon1 test = new Amazon1("amazon");
test.checkTheFrequency();
}
}
**
Best Regards
Kaushik Sur
On May 14, 3:00 pm, divya jain wrote:
> use binary tree and insert in it every character u come across. if the
> character is
