Sorry, ignore my post. Its totally broken.

2012/12/19 Andrey Ponomarev <[email protected]>:
> I think you can try the following. First, for each lowercase latin
> letter prepare an array of its indexes in the second string. Overall
> length is O(m) and the time is also O(m). Then, let the 'state' be two
> values for each lowercase latin letter: maximal subsequence length,
> ending with this letter and the leftmost position in the second string
> where this maximal subsequence ends. Initially set the state for each
> letter for (0,-1). Then, update the state for each prefix of the first
> string from prefix of length 1 to the whole string. At each step you
> have l_i - current letter in the first string, and state S. Try to
> update maximal length S[l_i] based on all possible last letters by
> binary searching array of indexes for allowed position of l_i with
> minimal index (i.e. leftmost). It gives you complexity O(n*c*log(m)).
> Where c stands for alphabet size.
>
> 2012/12/16 anupsingh <[email protected]>:
>> 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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