1st of all Happy New Year to All :)
Must be both have same length also test some other test cases, :)
Let me share things striking in mind , if you calculate the sum of power of
1st array by taking any number as base , remaining all the numbers as
exponent so if take 3 as base calculate
@top coder , if its given that that you have to nodes as input , 1st find
the two nodes , store their address then do any traversal to get the path
between them , inorder will good . i ams assuming you wants to print the
path between two such nodes.
hope it suffices :) let me know if approach
HI Anantha , I had similer discussion long time back , although its not
full final but similer ,i hope it will gives you some idea .
http://www.linkedin.com/groups/Detecting-Duplicate-Documents-1210167%2ES%2E55684470?qid=eb015784-36a3-498d-8441-558ace3b4038trk=group_items_see_more-0-b-ttl
@atul , its should be the either 1st even length palindrome or largest even
length palindrome in question , so here we go while making problem more
generic ,* given a string find the largest palindrome in that string *
So As I Coded it Some Times Back , Have a Look
you can use Suffix Tree for more better efficiency
Thanks
Shashank Mani
Computer Science
Birla Institute of Technology,Mesra
*http://shashank7s.blogspot.com*
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@atul ,FYI, another naive approach will to generate the all subset of given
set , thats the power set , has complexity O(n*2^n) , obviously not better
then upper one , but still you can try to figure out the sum problem , you
will get some relation to DP.
Thanks
Shashank Mani
Computer
You are given 'n' strings w1, w2, .., wn. Let Si denote the set of
strings formed by considering all unique substrings of the string wi. A
substring is defined as a contiguous sequence of one or more characters in
the string. More information on substrings can be found
@shashank : i am not sure if i am getting it right but are you saying save
address of 2 nodes.
now do inorder traversal considering node1 as the starting point , during
traversal if we find node2 then return.
if that so , then i dont think so it will work for all the cases.
please correct me if
@atul ,, why it won;t work , any clarification ?
Thanks
Shashank Mani
Computer Science
Birla Institute of Technology,Mesra
*http://shashank7s.blogspot.com*
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1) suppose node1 is at level = 4 of the tree and node2 is at level = 3 of
the tree
if you do inorder traversal from node1 i.e calling function inorder(node1);
now it will search form level 4 to the max depth of the tree
it will never reach node2 because it is at level = 3.
2) suppose root of the
@Shashank : as i have mentioned in the question , no sorting allowed.
if question would have allowed sorting then why not sort both array and
compare it would be much simpler and no need of doing costlier operation
like finding power.
complexity = O(nogn) + O(mlogm) + O(n)
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you are given with N points on graph. and a point A on graph and range
R you have to find the points that lies within the radius of R from
point A. with complexity less than O(N).
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Some very nice approaches have been presented but I still feels for
practical situations its better to sort and compare..All other
algorithms presented above restricts the max value for an element in the
array.In case the maximum value is small,we can simply count sort,and the
algorithm will
Assuming that the N points on the graph are represented in the form of
(x,y) i.e. cartesian coordinates..
Then say, A = ( a1, a2);
The equation of the circle centered at A would be
(x - a1)^2 + (y-a2)^2 = R^2...
Now, to find the points that lie within the range R, u need to check
the following
@ all
Ignore my previous post
On Jan 5, 5:58 pm, Lucifer sourabhd2...@gmail.com wrote:
Assuming that the N points on the graph are represented in the form of
(x,y) i.e. cartesian coordinates..
Then say, A = ( a1, a2);
The equation of the circle centered at A would be
(x - a1)^2 +
@atul..no need to sorting , complexity will be O(nlogn) only , sorting
makes searching easy so said to sort , u can linearly search that elemnt in
2nd array it will take O(m) in above case . finally complexity will be
O(nlogn) only
Let me know if anything wrong or missed something ?
Thanks
This problem is taken from www.codeforces.com.What can be the possible
approaches??
A smile house is created to raise the mood. It has *n* rooms. Some of the
rooms are connected by doors. For each two rooms (number *i*and *j*), which
are connected by a door, Petya knows their value *c**ij* —
@Saurabh DFS Should Work Here, Start from any room i , visit next room
connected to it .. then to this room so on , once you back track you
will back to the starting node ,this way you can find out .min.umber of
room he need to visit will be 2 times of traversal isn't it ?
posting the
depends on compiler i think..but most probably it compares the
addresses.
On Wed, Jan 4, 2012 at 12:20 PM, saurabh singh saurab...@gmail.com wrote:
@all.Your explanations work because probably all of you are using a
compiler that's behaving in the same way.Don't conclude from what
find the size of smallest cycle in the given graph ... tarjan should do the
trick.. dfs basically ... :)
On Thu, Jan 5, 2012 at 7:02 PM, WgpShashank shashank7andr...@gmail.comwrote:
@Saurabh DFS Should Work Here, Start from any room i , visit next room
connected to it .. then to this room
Yes I also initially thought soBut how do we take into consideration
the edge weights??The cycle can include such edges whose total cost may
come negative.
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Thu, Jan 5, 2012 at 10:01 PM, karthikeyan
Cut out the circle from the graph. The points on the cut-out circle
are the answer.
Don
On Jan 5, 6:17 am, dabbcomputers dabbcomput...@gmail.com wrote:
you are given with N points on graph. and a point A on graph and range
R you have to find the points that lies within the radius of R from
@Lucifier :
i was considering algo as mentioned by you and was considering heap with
indexes as mentioned in your first post i.e :-
/*
Also, though the MinH and MaxH is designed based on the element values
i.e A[i], but we will store only the index of the element i.e. 'i'.
// The above app will
The problem with the proposed depth first search is that it can try
many very long paths, requiring exponential time, before it ever finds
the correct cycle, even if the cycle is very short. A breadth-first
search will avoid this, and using dynamic programming principles can
accomplish the
@atul..
you are not getting it wrong.. Its absolutely correct
On Jan 5, 11:28 pm, atul anand atul.87fri...@gmail.com wrote:
@Lucifier :
i was considering algo as mentioned by you and was considering heap with
indexes as mentioned in your first post i.e :-
/*
Also, though the MinH
we can do bfs. from the point A do bfs until distance = R
On Fri, Jan 6, 2012 at 12:47 AM, shady sinv...@gmail.com wrote:
problem link ?
On Thu, Jan 5, 2012 at 5:47 PM, dabbcomputers dabbcomput...@gmail.comwrote:
you are given with N points on graph. and a point A on graph and range
R
@shady this problem is not from the online contest. i need this in my
project
@utkarsh please elloborate your idea in more detail
On Fri, Jan 6, 2012 at 9:40 AM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
we can do bfs. from the point A do bfs until distance = R
On Fri, Jan
i also think so data must be represent in some special form i heard
about K-D trees but not yet studied about it
On Fri, Jan 6, 2012 at 12:26 PM, Lucifer sourabhd2...@gmail.com wrote:
@all..
To decide which points are within the range R, we need to look a each
point..Hence the
the reason i asked you about problem link is because the solution to your
problem depends on the way you want to use it...
1. if each time there are new N points and radius R, and only one query
for it...then it just can't be O(n)
2. if N points are fixed and there are like 1+ queries then
@shady the N is same we just have to query the data again and again
On Fri, Jan 6, 2012 at 12:47 PM, shady sinv...@gmail.com wrote:
the reason i asked you about problem link is because the solution to your
problem depends on the way you want to use it...
1. if each time there are new N
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