int no_of_steps[arr_length] = {MAX};
if ( (arr_length==0) || (arr[0] == 0) ) //if there are no elements or
the very first element is 0 -> we can't jump anywhere
return MAX;
no_of_steps[0] = 0; //no jumps required to jump from element 0 to itself.
for (int i=1; i= (i - j) )//if it is po
^ cout<< no_of_steps[arr_length -1];
On Mon, Jul 9, 2012 at 8:44 PM, algo bard wrote:
> int no_of_steps[arr_length] = {MAX};
>
> if ( (arr_length==0) || (arr[0] == 0) ) //if there are no elements
> or the very first element is 0 -> we can't jump anywhere
> return MAX;
>
> no_of_steps[0] =
@ashish it wont be...first we r finding one end from any node dat is "r" n
den frm dat end we r traversing to other deepest end...
it is possible dat r b d intermediate node...n distance from r to v2 is
smaller than from r to v1
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farthest from
2: Find a vertex v1 | the farthest form r.
3: Find a vertex v2 | the farthest form v1.
won't v2 be farthest from r? or we are talking about alternate pats also
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Wed, Jun 20, 2012
As you traverse along and find the diameter of the tree, keep track of the
number of nodes thereby traversed. Let that be equal to n.
Now, centre is the node corresponding to floor((n+1)/2).
On Wed, Jun 20, 2012 at 5:19 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:
> I am asking again
I am asking again .. can any one please suggest better method for getting
the median on the longest path of the tree ..
efficient method .
On Tue, Jun 19, 2012 at 5:08 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:
>
> for getting diameter we can simply add the max height of left subtr
for getting diameter we can simply add the max height of left subtree and
max height of the right sub tree .
the main issue is how efficiently we find median on that longest path to
get the center of the tree .
can any bdy sugest optimized algo for that ?
On Mon, Jun 18, 2012 at 6:08 PM, rajesh p
I found it in some paper ;)
Diameter and center
De nition 4. The diameter of tree is the length of the longest path.
De nition 5. A center is a vertex v such that the longest path from v to a
leaf is minimal
over all vertices in the tree.Tree center(s) can be found using simple
algorithm.
Algor
I think this algorithm is used for calculating poset in graph.
On Sat, Jun 16, 2012 at 3:04 AM, Hemesh Singh wrote:
> + 1 for DK's solution. Is that a standard algorithm coz I feel like I have
> heard it somewhere ??
>
>
> On Mon, Aug 8, 2011 at 1:37 AM, DK wrote:
>
>> @KK: DFS and BFS are O(N)
+ 1 for DK's solution. Is that a standard algorithm coz I feel like I have
heard it somewhere ??
On Mon, Aug 8, 2011 at 1:37 AM, DK wrote:
> @KK: DFS and BFS are O(N) and Floyd Warshall is O(N^3).
> Could you please state how you can use the traversals directly to get the
> center? (And prove yo
@kunal patil:
U were proceeding in correct way...in next series u cud hav seen a
formation of arithmatico geometric series...
It doesnt matter what the value of no of faces in a dice is..ans will
be always 2...:)
My simplified soln: o+e+1
o=probability of odd number coming in 1
throw of dice
I
@Everyone: Wladimir has posted the correct solution to the problem and it is
an O(N) bottom up solution.
*The original solution:*
On Wednesday, 6 October 2010 21:10:40 UTC+5:30, wladimirufc wrote:
>
> Find the leaf of tree then put all leaf in a queue.
>
> While queue not empty:
> remove u fr
@kunal patil
expected number of rolls required to get even sum is inverse of that
i.e. 2 """
can we say like this all the time ?
i have understood the alternative method, thanks :D
On Sun, Aug 7, 2011 at 1:33 PM, Kunal Patil wrote:
> Probability of getting an even sum in one roll is 1/2..
>
@kunal patil:
how can u say "Probability of getting an even sum in one roll is 1/2."...
if tht is the case..can u find the ans in one go if the dice is having *5
faces*??
i mean the numbers on dice are 1 2 3 4 5 ...and it is unbiased...
what will be the *Probability of getting an even sum in one r
Probability of getting an even sum in one roll is 1/2..
Thus, expected number of rolls required to get even sum is inverse of that
i.e. 2.
Alternatively, Going by basics...
Let P(x) be probability of getting Even sum in x rolls.
P(1) = 1/2(Even)
P(2) = (1/2) * (1/2) (Odd + Odd)
P(3) = (1/2)
Traverse the tree inorder. Store the values in an array. Find the element in
the middle of the array.
On Sun, Aug 7, 2011 at 8:57 AM, subramania jeeva
wrote:
> 5
> /\
> 6 7
>/
> 8
>
> Here the centre is both 5 and 6 . we need to find both of them..:)
>
>
>
>
>
>
>
>
>
> Cheers
Dave shouldn't it be :: E(x) = Summation( xp(x) ) thus it should
be 1*1/2 + 2*(1/2 * 1/2) + 3*(1/2 * 1/2 * 1/2) .
On Sun, Aug 7, 2011 at 11:55 AM, Nitish Garg wrote:
> @Dave - Can you please explain how did you generate the series? Shouldn't
> it be : 1/2 + 2(1/4) + 3(1/8) and so
5
/\
6 7
/
8
Here the centre is both 5 and 6 . we need to find both of them..:)
Cheers
~ Jeeva ~
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Find the leaf of tree then put all leaf in a queue.
While queue not empty:
remove u from queue
remove u of tree
if some v adjacent a u become leaf
insert v in queue
the last vertice is the center of the tree
On Wed, Oct 6, 2010 at 8:08 AM, Ruturaj wrote:
> I am thinking
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