The concatenating of list(si) would have a size of O(mn) at worst case
and even the O(n logn) algorithm for solving the longest increasing
subsequence will not be fast enough.
Please tell me I am wrong. =w=
Parker
On 2012/12/17 9:28, Luís Fernando Dorelli de Abreu wrote:
Given the limits, th
Given the limits, those ideas are going to TLE I guess.
You can get an idea of a reduction that might work here :
http://www.cs.ucf.edu/courses/cap5510/fall2009/SeqAlign/LCS.efficient.pdf
2012/12/16 Faisal Dirie
> Use brute-force methods to devise an algorithm, then and then, use dynamic
> prog
Use brute-force methods to devise an algorithm, then and then, use dynamic
programming techniques to tweak your code to achieve an optimized one,
depending on what you want to do.
On Sun, Dec 16, 2012 at 6:20 PM, Faisal Dirie wrote:
> Check the links below.
>
> http://www.cc.gatech.edu/~ninamf/A
Check the links below.
http://www.cc.gatech.edu/~ninamf/Algos11/lectures/lect0311.pdf
http://en.wikibooks.org/wiki/Algorithm_implementation/Strings/Longest_common_subsequence
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hey can any one tell me the optimized method of solving LCS problem. I hv
optimized still it show TLE...
here is the link
www.spoj.com/problems/LCS0/
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Hi thefourtheye dIVi,
Basically your idea is correct and this problem can be solved using BFS.
However, it is not sufficient to use to solve this
problem because the answer to 19998 is
0 which is much larger than the
maximum number that could be represente
I am trying to solve http://www.spoj.com/problems/ONEZERO/
I referred so many internet posts about this, and they all talk about
storing reminders and building a tree. I am running a simple BFS, nothing
else... But this gets me WA :(
http://ideone.com/SZDn5T
Dont know whats wrong with this code.
Test cases of the problem are not testing this condition:
"he will not buy more than n packets of apples."
I got a success without putting check for this condition.
On Saturday, 15 December 2012 19:39:34 UTC+5:30, Marti wrote:
>
> I am having trouble figuring out the reccurrence for the follow
