the idea for O(|S|^2) is as follows....
Let string be s[1..n]
p[i][j]=1 if s[i...j] is a palindrome
p[i][j]=0 otherwise.
we can precalculate table p[i][j] as follows..in O(|s|^2)
p[i][i]= 1
p[i][i+1]=1 if s[i]==s[i+1], 0 otherwise
for all others..
p[i][j]= (s[i]==s[j] && p[i+1][j-1]==1).
now let m[i] represents the minimum number of palindromes needed to
represent s[1...i],
so m[n] is our required answer..
now...let us define m[i] recursively as follows..
m[0]=0, m[1]=1
m[i]=min { m[j-1]+1 where 1<=j<=i && p[j][i]==1}
what it actually says is that...we can go through all possible palindromes
which can be at the end of s[1...i],and check choosing which one is the best
option for us...
i guess its correct...if there is any problem with it then do suggest....
On Tue, Nov 16, 2010 at 7:56 PM, cheng lee <lichenga2...@gmail.com> wrote:
> how to do that in O(|s|^2)?
>
> 2010/11/16 anoop chaurasiya <anoopchaurasi...@gmail.com>
>
>> will O(|s|^2) dp work....or u need something more efficient than that....
>>
>>
>> On Tue, Nov 16, 2010 at 1:51 PM, lichenga2404 <lichenga2...@gmail.com>wrote:
>>
>>> A palindrome is a string which is identical to itself in reverse
>>> order. For example
>>> \ABAAABA" is a palindrome. Given a string, we'd like to break it up
>>> into the small-
>>> est possible number of palindromes. Of course, any one-character
>>> string is a palindrome
>>> so we can always break a length n string into n palindromes. Design an
>>> e cient al-
>>> gorithm to determine the minimum number of palindromes in a
>>> decomposition for a
>>> given string, and analyze the running time. You may assume you are
>>> given a proce-
>>> dure Palindrome(S) which runs in time O(jSj) and returns true if and
>>> only if S is a
>>> palindrome.
>>>
>>> How to solve this problem with the dynamic programming method?
>>>
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>>
>>
>> --
>> Anoop Chaurasiya
>> CSE (2008-2012)
>> NIT DGP
>>
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NIT DGP
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