@ ^
Why do you try hard not to understand the question or what one meant by the
question and instead try hard to find out flaws.
I mean ain't that obvious that you need to divide into minimum number of
groups?
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Isn't it just a coding question?
@ ^ : your code is not clean enough for anyone to read.
As far as data-structures are concerned... a stack and a queue suffice.
There is no algorithmic issue I can see. So, I guess you should solve these
problems yourself or some programming forum.
For those who
I don't see why you need O(n^2) time for rearranging.
It can be done in O(n log n) if you maintain the index along with every
element.
Then reordering would mean sort as per the indices.
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Let
No of flips reqd = solve(N, V)
L=left subtree.
R = right subtree
N = root node.
If V=0, then
only problem is when L = 1 and R=1. (otherwise atmax changing the root
node will do)
then ans = min(solve(L, 0), solve(R,0)) + (R=AND)?0:1
If V=1 then
only problem is when L=0 and R=0
Of course memoization is needed (Can be done by using an array and the fact
that it is complete binary tree.)
Also if the later half of the array is all 0, then 1 cannot be obtained at
the root and vice versa.
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@Divya : In that case, a greedy solution does not seem to exist.
You need to use the traditional DP way.
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Perhaps it is not available.
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For more
No special approach is needed. In O(log n), you can find the minimum element
of the array which makes your circular array - normal array.
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].Validity = d[n+1].Validity ) then d[n+1].Value
else d[i].Value
There is a practical issue that Validity may become larger than int etc...
however that too can be easily overcome.
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be rejected
straightforward from just this info.
Hence the complexity
T(m,n) = 2 * T(m/4,n/4) + c
Rohit Saraf
Third Year Undergraduate
Compter Science Engineering
IIT Bombay
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See it here : jfgi http://www.urbandictionary.com/define.php?term=jfgi
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well i just wanted u to know that there is a very nice place called
www.google,com where u can find many things.
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So, is this the place for that?
And apart from that, it is illegal to discuss about this on public threads.
(if you don't know, this thread is public)
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Why not just change the definition of when one number is bigger than another
and do normal sort ?
I guess that is better and simpler.
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Second Year Undergraduate,
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http
You must be doing some useless iterations . Otherwise, TLE is too strange
for this prob.
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http://www.cse.iitb.ac.in/~rohitfeb14
On Sat, Jun 19, 2010 at 9
It was my lab assignment prob last year. Will send u if i happen to find it
by chance.
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On Mon, Jun 14, 2010 at 10
No
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On Sun, Jun 20, 2010 at 3:22 PM, Shravan shravann1...@gmail.com wrote:
Did you see the code which I have
You need to specify the max level upto which you want it to work.
For rest, is there any prob?
I can't see any.
you see ninety - 90
u see thousand - 90*1000
+
and so on
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Second Year Undergraduate,
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You start from a vertex, then if you reach that vertex before dfs on that
vertex finishes, then the edge u used to reach it is a back edge.
And if you reach after the dfs finishes on that node, it's a cross edge.
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Someone ban this spammer
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Yes right, i forgot the 1
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On Thu, Jun 17, 2010 at 6:05 PM, Dave dave_and_da...@juno.com wrote:
No. The greedy
If the coins are all multiple of some number k, you can greedily give
as much as possible to the higher domination. Otherwise still, there
is an optimal substructure and u can make a recurrence and use
memoization(i.e. DP)
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@Dave: The greedy will only work if the coins are k,2k,3k,4k, nk without
any of these missing
Clear?
(Perhaps i did not write it clearly as i was on mobile)
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@vivek : was that a joke?
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On Mon, Jun 14, 2010 at 11:40 AM, souravsain souravs...@gmail.com wrote:
@sharad: Do
Just to point out :
how many times have you all repeated this --
Xor works only -- even number of times. It will not
work...
Why don't you all read some earlier posts before posting. :P
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I read that. But still it should not be compiled as per the standard.
The latest GNU C/C++ compiler correctly fails to compile this
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http
@divya: u r rite.. that * should not be there
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On Sun, Jun 13, 2010 at 11:07 AM, divya jain sweetdivya
which compiler do you use?
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On Sun, Jun 13, 2010 at 10:46 AM, divya jain sweetdivya@gmail.comwrote:
hmm
what is this for...
and which conversion are you talking abt?
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On Sat, Jun 12, 2010 at 11:20 PM, divya sweetdivya
M-speed is private and you cannot call it from outside myPlugin. (though i
did not understand what u wanted to say)
Can you write ur prob explicitly?
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http
Make those functions public
And even if M_speed is public of another class, it is still it another
class, you cannot just address it like m_speed in other classes.
Does it help?
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Rohit Saraf
Second Year Undergraduate,
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yes.. it should... make sure your virtual function is either public or
protected but not private.
and if it doesn't can u tell me the error?
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Put the function speed or its declaration inside the class b. Did u forget that?
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On Sun, Jun 13, 2010 at 1:47 PM, divya jain sweetdivya@gmail.comwrote:
i use tc
On 13 June 2010 13:11
@jalaj: exactly...
so you(@divya) are right. Sharad's ans was right but logic wasn't.
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On Sun, Jun 13, 2010 at 2:35
If you are not able to print the long int and that's the prob, you can use
%ld instead of %d
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On Sun, Jun 13, 2010
@Souravsain : Is there any serious problem in this. Anyone can just add a
[C++] in the subject and uninterested people can make filters in gmail :)
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http
I agree mass bombarding with such questions is not very good.. but one
doesn't join groups and all for getting a few doubts cleared.
Anyways, i have no problem with anything. :D
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In any directed graph just check if dfs has a back edge. For
undirected, check if there is a nontree edge
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Repeated q. Search in the group
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@ram : i guess you have used some longer string and not strings
btw.. what is Mingw ?
gcc/g++ is not mingw, i guess
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when assigning to type ‘char[20]’ from
type ‘char *’
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On Sun, Jun 13, 2010 at 8:13 AM, Mahesh_JNU mahesh.jnumc
@Satya: I don't think what you have coded will work.. though i have not read
the whole code.
Don't you think a simple divide and conquer scheme would work...(almost like
the mergesort)
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a (
()
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On Fri, Jun 11, 2010 at 12:07 PM, Vivek Sundararajan
s.vivek.ra...@gmail.com wrote:
@Rohit
Consider : (a+)b
The above is not well formed! :)
On 11 June 2010 11:58
Fibonacci numbers can be calculated very efficiently
using matrix multiplications.
I hope you can think it/google it now.. that is better than me writing so
much again :)
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Second Year Undergraduate,
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I guess it can be done in efficiently using a simple divide and conquer
scheme almost imitating mergesort.
Can you think of it now? :D
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http
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On Thu, Jun 10, 2010 at 11:47 AM, Sundeep Singh singh.sund...@gmail.comwrote:
@rohit: fibonacci sequence may
) = -1
( = +1
Keep a sum of all these as u iterate.
That should never be negative
Plus check for these types (if you need correct arithmetic expressions as
well
some operator followed by )
* or / after a (
()
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c
1
6,2
u might be expecting 5,1 if u are forgetting the newline character :)
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@debajyoti: read the prob before posting
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On Wed, Jun 9, 2010 at 2:37 PM, Debajyoti Sarma
sarma.debajy
I had answered this question(of multiple lists) 2 or three days back.
Go into the archive if u wanna see :P
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On Wed
@junta : are fibonacci sequence is the answer of the prob, it is not used :D
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On Wed, Jun 9, 2010 at 9:13 PM, Rohit
@senthilnathan : very nice indeed
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Your time complexity is not O(c) but O(n^2)
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Fib comes because she wants the number of such sequences
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which is just the recursive version of the abovementioned iterative
solution.
P.S. -Please remove this quoted text when you are composing
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So u want efficient algo for fibonacci numbers?
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better.
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getting fibonacci nos is trivial using matrix multiplication in almost
constant time.
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Of course you should do it via swappings.. why would one think of anything
else :)
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On Mon, Jun 7, 2010 at 10:39 PM
but actually we need something better as per prob,
cannot be done in O(n).
so we need to think of something like O(n lg n) or O( n ^ 3/2) :)
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http
No it will be O(n).
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Algorithm
Tokenization is done in linear time. Just save the words in an list (And
what makes you think of non-linearity in tokenization!)
And then iteration over the tokens is trivially linear.
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Actually he means the case when you implement quicksort using stacks.
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.
Dynamic programming is a more general approach which is generally used when
there is optimal substructure and has mostly found use in doing exponential
looking problems in polynomial time.
i hope u understood
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If you make it unsigned short int.. then it goes to 65535 on g++/gcc
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Ok. so you have a list.
Iterating over it is linear isn't it?
Ahh... you will need a doubly linked list or an arraylist.does it solve ur
prob?
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oh... why did u put %u.
i did not even notice that :P
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What makes you think it is O(n^3).
I did not read the code one but divya's solution seems to be O(n^2)
for worst case.
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Just read your code. It wont even work.
Do you assume only one even length palindrome!?
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Can u elaborate on the 2nd step.
btw, did u understand my soln?
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This is almost the most basic dp. Read some of the examples from eva
tardos. That would help.
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@jitendra: How can you say that no polynomial algo can be found. I think you
are wrong.
A simple memoized formulation will make this polynomial because of the
optimal substructure..
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Oh i am talking only about printing the total number of solutions...
If you backtrack... Obviously it would not be polynomial
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if you short the larger one first, then the smaller one will be in the
stack for a long time. Which is wasting stack space.
Now stop shorting and start sorting ! :)
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Make a hashmap... Any problem doing that?
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What precisely did you not understand??
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Have you posted the same question twice or i am feeling sleepy?
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Exactly that's what you need to do.
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@divya: but still haven't you seen Jagadish's example? It is a
counterexample to your greedy approach.
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Still the solution will be similar and actually a bit simpler.
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A*(B+C*D)
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So there is a prob algoose A*(B*C) and a*(b*c+d)
i hope you understood
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why don't you make use of the optimal substructure.
You can easily use the recurrence relation as DP
@all : people don't just paste your code. Words are better than code
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!
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+200)-500)abs((2000)-(500+200))
so we ll put 200 in array B. now since B has n/2 elements rest of the
element goes to array which is 1.
so the ans is
A={2000,1}
b={500,200}
On 15 May 2010 19:10, Rohit Saraf rohit.kumar.sa...@gmail.com wrote:
so what will ur algo give for array
}
the difference is 40
the correct ans:
A = { 110, 100 , 10 }
B = { 90 , 70 , 60 }
the difference is 0
i don't believe a greedy algorithm would work for this problem!
On Mon, May 17, 2010 at 5:06 PM, Rohit Saraf
rohit.kumar.sa...@gmail.comwrote:
@divya : descending order sorting works
@Navin: and that works ! :)
@all : i am sure no heuristic/greedy strategy can be applied.
@divya : did you check your array partitioning algorithm with my example !
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suggesting will be polynomial.
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@anurag : won't work
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Read it up : http://www.eecs.umich.edu/~qstout/pap/CACM86.pdf
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@Nikhil : For that size of subtrees at all nodes needs to be maintained. And
in that case this is a trivial problem.
@Sathaiah : See my solution :)
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/2 where n is
the no. of elements in the given aaray. then store rest element in the
other array.
5. repeat step 5 until both array A n B get n/2 elements..
hope my approach is clear and correct.
comments are welcomed.
On 15 May 2010 08:47, Rohit Saraf rohit.kumar.sa...@gmail.com wrote
.
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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A simple DP should work. Should it not?
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
On Fri, May 14, 2010 at 4:15 PM, Sathaiah Dontula don.sat
I think... it will work :)
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
On Fri, May 14, 2010 at 2:20 PM, divya sweetdivya@gmail.com wrote:
Given a graph's
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