Jordan Greenberg posted:
"Using a spell checker is probably a good idea as well."
Having some meaninful content in your post, is an even better idea,
Jordan. This is NOT a spelling bee forum!
Adak
.
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Gene wrote:
>
> #include
>
> void interleave(char *a, int ia, char *b, int ib, char *r, int ir)
> {
> if (ir < 0)
> printf("%s\n", r);
> else {
>if (ia >= 0) {
> r[ir] = a[ia];
> interleave(a, ia - 1, b, ib, r, ir - 1);
>}
>if (ib >= 0) {
> r[ir] = b[ib];
>
But to setup a threaded binary tree needs to traval the whole tree first.On 5/13/06, thomas <[EMAIL PROTECTED]
> wrote:
Feng wrote:
When you visit a node N which has two children ( binary
tree, for example ) L(N) and R(N), if the next step is visiting L(N),
then save L(N) and make L(N) p
Feng wrote:
When you visit a node N which has two children ( binary
tree, for example ) L(N) and R(N), if the next step is visiting L(N),
then save L(N) and make L(N) points to N's father which has been saved
before. The pointers to fathers take place of stack or recursion.
On 5/13/06,
Maybe that is why it continues to grow despite all the hard work to
stop it:
http://fraudwar.blogspot.com/2006/05/are-we-addressing-cyber-crime-from.html
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"Algorithm G
When you visit a node N which has two children ( binary tree, for example ) L(N) and R(N), if the next step is visiting L(N), then save L(N) and make L(N) points to N's father which has been saved before. The pointers to fathers take place of stack or recursion.
On 5/13/06, Manu <[EMAIL PROTECTED]>
Imran Mohammed Abdul Muqsith wrote:
> Hi,
>
> Can anyone give the algo to generate all possible interleavings of two
> arrays (better if its in C or Java)
>
> E.g.
> A1[] = {a,b};
> B1[] = {c,d};
>
> # Results = (2+2)!/(2! .2!) = 6
>
> abcd
> acbd
> acdb
> cabd
> cadb
> cdab
#include
void inter
Hi,
Can anyone give the algo to generate all possible interleavings of two arrays (better if its in C or Java)
E.g. A1[] = {a,b};B1[] = {c,d};
# Results = (2+2)!/(2! .2!) = 6
abcdacbdacdbcabdcadbcdab
-- Mohammed Abdul Muqsith ImranDepartment of Computer Science and Engineering,Indian Institute o
Find an iterative method to do the traversal of a tree (inorder
say)
without using recursion and without using stack...
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To post to thi
well if the index of the elements are not imp then it's fine...but if
they are then may be we will have to use some other way..
like find all the indexes from left (i) and right(j) till the sum is
equal to the required sum and then find all possible pair of the
indexes.
thnx
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well the set (X + Y) will have O(n^2) elements, where X and Y have n
elements.
As an example
X = {1, 2}
Y = {3, 5}
X + Y = {1 + 3, 1 + 5, 2 + 3, 2 + 5} = {4, 5, 6, 7}
Now the problem is to find the median of X + Y in subquadratic time.
thx
aj
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Why do you want both the pairs to be printed? I guess the problem's
originality is only in finding unique pairs summing to a particular
number, so there's no point in printing repeating pairs.
Besides, even if you want to do that, daizi's code can be altered quite
simply to print repeating pairs.
i m referring to daizi's algo.
n = size of num
i = 0
j = n-1
while(i < j) {
c = num[i] + num[j]
if(c < given number) i++
else if(c > given number) j--
else show and j--
}
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well i guess the algo is fine if the elements are not repeating but
when they repeat i think there is some problem as all possible pairs
are not printed
for eg. 1,1,2,3,4,5
then this algo prints <1,5> only once
becoz then poitner j decrements and we never compare the next 1 with
5
-
aj wrote:
> How about finding the median of (X + Y) in subquadratic time.
> X + Y = {z | z = xi + yj for some i and j}.
>
> I could only reduce the space complexity from O(n^2) to O(n) but time
> complexity still
> remains O(n^2).
>
> thx
> Aj
Can't you merge the two sorted array in O(n) into a
W Karas wrote:
> Vibhu wrote:
> > How wud you handle the case where
> >
> > X= {1, 3, 5, 7, 9}
> > Y= {2,4,6,8}
> >
> > What wud be timecomplexity ?
>
> Here is a more general solution, covering cases
> like your example where X and Y have different
> dimensions:
>
> Let Nx be dimension of X, Ny
On 5/12/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> [...] I Spend My Whole Summer Devoted
> To Programming And Learning...
please, please, please, you are 13, go outside,
play football with your friends, try to go by bike, go
to the seaside, swim, play, and enjoy your being 13,
you have an
How about finding the median of (X + Y) in subquadratic time.
X + Y = {z | z = xi + yj for some i and j}.
I could only reduce the space complexity from O(n^2) to O(n) but time
complexity still
remains O(n^2).
thx
Aj
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