@Rahul : if you check the flow properly ,(lets concentrate on the root node
, call other as left and right subtree) you will find that after done with
reversing root->left->left subtree , it reaches root(backtrack to root)
node and then swap root->left and root->right.
now because it is inorder way
@rspr: Are you talking about dependencies in my approach? or adding new
constraints?
In my DP approach, f(n,m) only depends on f(n-1,m'), f(n-2, m') ,...,
f(1,m') (m' <= m)
Using s(n,m) denoting the number of all graphs of n vertices and m
edges, obviously s(n,m) = C(n(n-1)/2, m)
Instead of
What about just the root being reversed?
Why is it different only in case of root?
Rahul
On Feb 9, 10:52 pm, Don wrote:
> Because it reverses one side twice and the other side not at all.
> It does a lot of work to accomplish nothing.
> Don
>
> On Feb 9, 9:06 am, Rahul Menon wrote:
>
>
>
>
>
>
Hi All,
Anyway to implement this in a good manner
http://www.spoj.pl/ARHN/problems/PRINCESS, solution is simple,
check for ones and then generate numbers is an increasing number...
so if a number n = 22(10110)
then for k = 1, ans = 2(10)
then for k = 2, ans = 4(100)
then for k = 3, ans = 6(110)
S
well i have used three recurrences :P formed them by following a
traditional approach
f[i] = f[i-1] + 2*g[i-1] + h[i-1] + f[i-2];
g[i] = f[i-1] + g[i-1];
h[i] = f[i-1] + h[i-2];
On Thu, Feb 9, 2012 at 7:19 PM, Kunal Patil wrote:
> I am solving spoj problem Tiling a Grid
It can't be done in O(log n) unless the array is sorted or there is
some other special property, for example, if all values x are known to
be contiguous, allowing you to use a binary search to find the first
and last location of x.
In the general case it is impossible in O(log n) because you have
Because it reverses one side twice and the other side not at all.
It does a lot of work to accomplish nothing.
Don
On Feb 9, 9:06 am, Rahul Menon wrote:
> How come it just reversed the root? I still dont get it!
>
> Rahul
>
> On Feb 9, 7:57 pm, Don wrote:
>
>
>
> > It appears to be an attempt to
How come it just reversed the root? I still dont get it!
Rahul
On Feb 9, 7:57 pm, Don wrote:
> It appears to be an attempt to reverse the tree. However, there is a
> problem. It reverses the left sub-tree, then swaps the left and right
> sub-trees. Then it reverses the right sub-tree. But wait!
Thanks!
I knew that it wont reverse the tree but was not sure about how it
reversed just the root.
On Feb 9, 7:57 pm, Don wrote:
> It appears to be an attempt to reverse the tree. However, there is a
> problem. It reverses the left sub-tree, then swaps the left and right
> sub-trees. Then it rev
It appears to be an attempt to reverse the tree. However, there is a
problem. It reverses the left sub-tree, then swaps the left and right
sub-trees. Then it reverses the right sub-tree. But wait! The original
left sub-tree which we reversed is now the right sub-tree, so we
actually unreversed it.
What does this function do?
void function(Node **node){
if(*node!=NULL){
function(&(*node)->Left);
Node *temp;
temp = (*node)->Left;
(*node)->Left= (*node)->Right;
(*node)->Right = temp;
functio
I think all the answers are wrong. Also pre-order is probably the
wrong term here. The conventional term is prefix or Polish notation.
You'd break up this expression at the level of lowest precedence as:
(A - B) - C
where A = ~16, B = ~14 / ~12, and C = 2 * 8 .
Note I'm using ~ for unary nega
I am solving spoj problem Tiling a Grid With
Dominoes.(http://www.spoj.pl/problems/GNY07H/)..
I am not able to come up with a recurrence relation..
One of my friend said it has the recurrence relation as f(n) = f(n-1)
+ 5*f(n-2) + f(n-3) - f(n-4).
I am not convinced and have trouble deriving this f
>From the following options, select the correct pre-order
representation of
the following expression.
– 16 – – 14 / – 12 - 2 * 8
Please do answer how you arrived at the answer!
Answers
-–16--/14–12*28
-–16--/-1412*28
-–16-/–14–12*28
*/--–16-14–1228
--
You received this message because you are s
@Terence: The DP approach is nice.
if we have constraint likewhile choosing the first 3 edges if it
makes a triangle so we will require n edges to connect the graph
completely...in the same fashion...after selecting 2 more
edgesagain we have to check that is it making more
trianglethen
i dont think its possible for unsorted array in O(logn)..
Regards,
PAYAL GUPTA ,
NIT-B
On Thu, Feb 9, 2012 at 6:02 PM, atul anand wrote:
> @payal : this will work for only sorted array not for unsorted.
>
>
> On Thu, Feb 9, 2012 at 5:50 PM, payal gupta wrote:
>
>> http://www.geeksforgeeks
@payal : this will work for only sorted array not for unsorted.
On Thu, Feb 9, 2012 at 5:50 PM, payal gupta wrote:
> http://www.geeksforgeeks.org/archives/4722
> O(logn) soln...
>
> Regards,
> PAYAL GUPTA,
> NIT-B
>
>
> On Thu, Feb 9, 2012 at 5:33 PM, atul anand wrote:
>
>> ignore my commen
http://www.geeksforgeeks.org/archives/4722
O(logn) soln...
Regards,
PAYAL GUPTA,
NIT-B
On Thu, Feb 9, 2012 at 5:33 PM, atul anand wrote:
> ignore my comment...
>
>
> On Thu, Feb 9, 2012 at 4:15 PM, Ashish Goel wrote:
>
>> log n is impossible. the other solution i thought was of building a tree
ignore my comment...
On Thu, Feb 9, 2012 at 4:15 PM, Ashish Goel wrote:
> log n is impossible. the other solution i thought was of building a tree
> where each node contains value and its count. and then building a heap out
> of these counts, but this will be overkill.
>
> the fact that rest of
log n is impossible. the other solution i thought was of building a tree
where each node contains value and its count. and then building a heap out
of these counts, but this will be overkill.
the fact that rest of the n/2 elements are not unique is the killer in the
logic otherwise only n/2+1 elem
i guess can be done using binary indexed tree.
On Thu, Feb 9, 2012 at 2:07 PM, Prakhar Jain wrote:
> But in this post, we don't have prior information about what can be
> possible majority element.
>
> According to my question, we know that either x is the majority element or
> there is no major
But in this post, we don't have prior information about what can be
possible majority element.
According to my question, we know that either x is the majority element or
there is no majority element.
Can we use that information to reduce complexity to O(log n)..???
--
You received this message b
Thanx for reply
can you provide some code snippet here?
or some link for it.
On Tue, Feb 7, 2012 at 11:30 PM, Prem Krishna Chettri wrote:
> This is a simple implementation to Factory Design Pattern. What you have
> to do is make an arbitrary class (Your Adapter) and always call this.
> Howeve
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