@Gene : your code will work fine by changing the argument passed from
main(), you just need to call rite f(n, 1, 1); from main instead of f(n,
1, 0);
On Mon, Mar 19, 2012 at 10:10 AM, atul anand atul.87fri...@gmail.comwrote:
@all : i guess question is on Fibonacci coding.
here you can find
If anyone have any idea please share .
On Mar 15, 12:06 pm, Manjay kumar manjay.bit...@gmail.com wrote:
DocumentFormat.OpenXml library of .Net
On Thu, Mar 15, 2012 at 12:35 PM, Manjay manjay.bit...@gmail.com wrote:
From where can I have Knowledge of this library of .NET.
If anyone have any
hii supraja can u mail me the link for your blog plz.
On Sun, Mar 18, 2012 at 6:38 PM, Supraja Jayakumar suprajasank...@gmail.com
wrote:
Hi
Others are also welcome to comment on the code. If links are allowed in
algogeeks, I might send my wordpress blog link that explains this problem
in
Thanks.
I noticed this too. If the n'th 1/0 digit is supposed to correspond
with the n'th fibonacci number, then my original code would have been
right. But the example isn't done this way.
I fixed the code to match the example the evening of the 18th
(Eastern time), but I guess the change is
verticalSum( node * root, List * sum)
{
if(root == null) return;
sum-data += root-data;
verticalSum(root-left, sum-left);
verticalSum(root-right, sum-right);
}
1
/ \
2 3
/ \/ \
4 5 6 7
a b c d e
Vsum(1, c) c= 1
Vsum (2,
@gene it does show your updated code.
@atul from the given input it seems different from Fibonacci encoding.
On Mon, Mar 19, 2012 at 5:32 PM, Gene gene.ress...@gmail.com wrote:
Thanks.
I noticed this too. If the n'th 1/0 digit is supposed to correspond
with the n'th fibonacci number, then
nice explanation aman and prashant
1(0) / \ 2(-1) 3(1) / \ / \ 4(-2) 5(0) 6(0) 7(2)
As you see this example, each node has an extra attribute(not necessary
though) which tells its distance from the root node. Take map and as you
traverse the tree in any order, add the count to the map value.
anything that can help people learn is always allowed. :)
On Sun, Mar 18, 2012 at 6:38 PM, Supraja Jayakumar suprajasank...@gmail.com
wrote:
Hi
Others are also welcome to comment on the code. If links are allowed in
algogeeks, I might send my wordpress blog link that explains this problem
@Gene : yeah i skipped ur updated code...its dere
@shady : it is similar to fib encoding...you just need to reverse the
output from gene code and append '1' at the end...
while decoding it ...this extra '1' is not considered..
i am nt sure but maybe the reason for adding '1' at the end is to
@supraja ..can u give example..code not needed..
@all..plz post me example.i dnt know what is vertical sum..i wana know only
that..thnx...
On Mon, Mar 19, 2012 at 7:31 PM, shady sinv...@gmail.com wrote:
anything that can help people learn is always allowed. :)
On Sun, Mar 18, 2012 at 6:38
if tree is like
1 / \ 2 3 / \ / \ 4 5 6 7
/ \
12 -8
then vertical sums are
12(1 + 5 + 6)
2
4
-6(2+-8)
3
7
12
On Mon, Mar 19, 2012 at 9:05 PM, rahul sharma rahul23111...@gmail.comwrote:
@supraja ..can u give example..code not needed..
@all..plz post me example.i dnt know what is vertical
oops no 2 there
On Mon, Mar 19, 2012 at 9:36 PM, shady sinv...@gmail.com wrote:
if tree is like
1 / \ 2 3 / \ / \ 4 5 6 7
/ \
12 -8
then vertical sums are
12(1 + 5 + 6)
2
4
-6(2+-8)
3
7
12
On Mon, Mar 19, 2012 at 9:05 PM, rahul sharma rahul23111...@gmail.comwrote:
@supraja
This was a MS question asked recently on Run length Decoding. I was
given
Input- a3b5c3d2
And the output should be ddcccbaaa
Assuming that the memory given is sufficient to accomodate the whole
string.
And this conversion should be inplace. ie the output string should not
use another array.
keep a pointer and just write the count with the corresponding character on
the same character array.
On Mon, Mar 19, 2012 at 10:38 PM, ATul SIngh atulsingh7...@gmail.comwrote:
This was a MS question asked recently on Run length Decoding. I was
given
Input- a3b5c3d2
And the output should be
the catch here is how you represent aa?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Mar 20, 2012 at 1:56 AM, shady sinv...@gmail.com wrote:
keep a pointer and just write the count with the corresponding character
on the same
It's not hard if all the run lengths are at least 2.
void decode(char *buf, int in_size, int buf_size)
{
int i, j, rl, p;
char t;
// Reverse the input.
for (i = 0, j = in_size - 1; i j; i++, j--) {
t = buf[i]; buf[i] = buf[j]; buf[j] = t;
}
// Copy to end of buffer (carefully)
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