@Dave :
Wonderful answer sir..
On 31 October 2011 12:50, Dave wrote:
> @Ankuj: Your method would be O(n), where n is the number of nodes in
> the tree. However, it can be done with a sequence of tree rotations.
> If the desired node is not the root of the tree, rotate it with its
> parent. This
from where does the index starts, 0 or 1 ? in this, array to be moved is
{7, 5, 8} ?
and source array the destination
| |
{9, 7, 5, 8, 1, 5, 4, 8, 10, 1}
please explain move_set o
:)
On Tue, Nov 1, 2011 at 12:54 AM, WgpShashank wrote:
> @Anup You Seems to Active Member , u remembered that :)
>
> +1 to You.
>
> Thanks
> Shashank
>
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No constraint on path. Though it is not necessary that it starts from
root only
On Oct 31, 10:02 pm, Prakash D wrote:
> any constraints for path?
>
>
>
>
>
>
>
> On Mon, Oct 31, 2011 at 11:19 AM, Ankuj Gupta wrote:
> > Print all path of the tree that sums up to the given value. The path
> > may
@Ankuj: Your method would be O(n), where n is the number of nodes in
the tree. However, it can be done with a sequence of tree rotations.
If the desired node is not the root of the tree, rotate it with its
parent. This preserves the BST property and makes the desired node one
level closer to the ro
@Anup You Seems to Active Member , u remembered that :)
+1 to You.
Thanks
Shashank
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@shashank , i agree.
To reduce this overhead , we can implement prefix trie in bit-form or DAWG
or lots of compression techniques are there at the cost of complex coding.
On Tue, Nov 1, 2011 at 12:35 AM, WgpShashank wrote:
> @Mohit Space Overhead Will be more if m,n are large isn't it ?
>
>
> S
Search For Bin Packing Problem on Wiki .
Thanks
Shashank
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My Dear friend ,
For Getting the Path above the source node here is the simple step ..
The below table illustrates it :-
During traversing the in BFS way from the root until we find the
source node .
Current Node = a | b c | d e
Distance from Root = 0 | 1 1 | 2 2
PrevNode
@Mohit Space Overhead Will be more if m,n are large isn't it ?
Shashank
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structure of n-ary tree
Class:Node
String:label
List children
Method
List TreeToLinkedList(Node n)
{
return list;
}
return List of all nodes;
[image: image.png]Pics Taken From Google Image .
N-Ary Tree
Output head-> 1-2-3-4-5-6-7-8-9
I had a doubt , if we simply go on constructing the tree from preorder
traversal one gets the same tree back am I missing something here
On Oct 19, 7:49 am, sravanreddy001 wrote:
> @Sunny, Rahul:
> Thanks a lot.. :)
>
> @Anshu: the code is perfect, This would be h = n* (height of BST) --> O
One (compressed-optional) prefix trie with each course ptrs to students and
another prefix tree for students having ptrs to courses. Guaranteed O(m)
search time instead of O(n*m) if using hash table where m avg size of word
at hand to be compared to n nodes.
On Sun, Oct 30, 2011 at 7:31 PM, WgpSha
Any body got any idea of just how to approach It need a DP algo.
On 10/30/11, SAMMM wrote:
> Suppose u have a square matrix, where every cell is filled with 0 or
> 1 . U need to find the maximum subsquare such that all four borders
> are filled with all 1s.
>
> Ex:-
>
> 1 0 0 1 1 0
> 1 0 1 1
thnx all
On Mon, Oct 31, 2011 at 10:13 PM, Vandana Bachani wrote:
> Hi Mohit,
> Bellman-Ford algorithm is a dynamic programming algorithm but u need it
> only if dijkstra's SP wont solve the problem... and Dijkstra's algo works
> only for +ve weights. So if u r sure that there maybe negative weig
This is like the TOPOLOGY SORT .
Repeat the steps until all the elements of orginal set is visited:-
Do
First need to add all those elements in a set whose inorder is Zero .
Then create another set whose inorder is 1 , after making set 1 ,
then decrease the inorder by 1 for those elements who was
any constraints for path?
On Mon, Oct 31, 2011 at 11:19 AM, Ankuj Gupta wrote:
> Print all path of the tree that sums up to the given value. The path
> may start from any node.
>
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+1 abhishek
On Mon, Oct 31, 2011 at 6:26 PM, abhishek kumar wrote:
> Represent the dependencies as a graph. Store all the values in a list. For
> each vertex in the graph find all values for which there is no edge from
> the vertex. If these values are there in the list, remove them from the
> l
How to convert a Binary tree to BST ? Naive way is to create each node
of Binary tree one by one and keep on creating the BST.
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Given a node of a BST, modify it in such a way that the given node
becomes the root. The tree should still be BST. One way I could get is
store the Inoder traversal of the tree. Find that node in the
traversal and recursively make the BST.
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Hi Mohit,
Bellman-Ford algorithm is a dynamic programming algorithm but u need it
only if dijkstra's SP wont solve the problem... and Dijkstra's algo works
only for +ve weights. So if u r sure that there maybe negative weights then
you will need to use bellmann ford which is a DP algo.
On Mon, Oct
I am not getting what you said. For given tree
a(2)b(7)c(5)d(2)e(6)f(9)g(5)h(11)i(4) if i say given node is 6 and
distance is 1 then the output should be 7,5,11.
The nodes below the given nodes can be easily printed. I am not
getting the way to print nodes above the given node
On Oct 31, 1:30
There are cycles in recursion,there is no recursion tree over here, which
can be traversed in bottom up fashion,so no DP.
Shortest path algorithms will be the answer
On Mon, Oct 31, 2011 at 9:30 PM, Don wrote:
> Use a path cost matrix of the same size as the given matrix. Set all
> locations to
Use a path cost matrix of the same size as the given matrix. Set all
locations to a maximum cost value.
Use a queue to store locations which still need to be processed.
Start by setting the cost to the destination equal to the cost of that
location and adding that location to the queue.
Then, while
yeah , my algo wont work for these cases :(.
On Mon, Oct 31, 2011 at 6:32 PM, sunny agrawal wrote:
> @Mohit
> that will also not work
> example: {1,1,1,2,2,2,3,3,3}
>
> i think all the methods that try to fill the matrix(considering k sets as
> k rows) either horizontally or vertically will fail
Represent the dependencies as a graph. Store all the values in a list. For
each vertex in the graph find all values for which there is no edge from
the vertex. If these values are there in the list, remove them from the
list and create a set of the vertex and the removed values.
If the values are n
@Mohit
that will also not work
example: {1,1,1,2,2,2,3,3,3}
i think all the methods that try to fill the matrix(considering k sets as k
rows) either horizontally or vertically will fail
we need to fill these both horizontally and vertically at the same time
depending upon the frequency of elements
Hi SAMM,
The above code is not clear enough to understand . Sorry for that.
My idea was , like for above example : map will contain frequency of all
distinct elements.
so freq[1] = 6
freq[3]= 1
freq[5]=1
freq[6]=1
Now for n=9 and k=3
P1={1,3,5};
and now after this partition freque
I knew this could be done using Min Path finding algo. But what about DP
for this problem , guys?
On Mon, Oct 31, 2011 at 3:49 AM, SAMM wrote:
> This can be done using the Dijkstra's algorithm , Start frm the source
> frm the Destination (In this example from (2 2)) . You need to
> consider the
Hey guys ,
I am trying to solve this DP problem :
http://community.topcoder.com/stat?c=problem_statement&pm=10033&rd=13513
SRM 422, DIV 2 ,level 2.
Here is my DP solution. But it is not working. Can someone please tell me
the flaw in this code :
#include"topheader.h"
int prime(int i)
{
retur
Given a set of values..in which there are some dependencies..
Eg.. x y z a b
Dependencies: x,y
a,z
Note that dependency is not transitive..Is it possible to separate
these elements into sets such that no two elements in the same set are
dependant and we should end up with the least number of se
As the problem states that the distance can be upwards and downwards .
So we considering both the case . I am going to implement BFS to
implement it (Because requires the poped elements to trace back to the
Source node to check whether path is sorted)
Consider the tree Given in this link .. I am
Print all path of the tree that sums up to the given value. The path
may start from any node.
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