T[i][j] = 0 (i < 0 || j >=n || i >= j || s[i] != s[j])

T[i][j] = 1 + T[i-1][j+1]




On Fri, Jun 6, 2014 at 12:19 AM, kumar raja <[email protected]>
wrote:

> If i get u correctly, this is the recurrence relation ( i feel this makes
> many people to understand the solution rather than looking at the code)
>
>
> T[i..j] indicates the length of longest even length palin string ( not
> exactly palin but i think u know what i am saying) with i as begin and j as
> ending point.
>
> T[i,j] = 0 if i>=j
>            T[i+1,j-1]+1 if s[i]==s[j]
>           0    else
>
> And u find max value in the table which is the answer
>
> So time complexity  = space complexity = O(n^2).
>
> Correct me if i am wrong
>
>
>
> On 5 June 2014 23:44, Saurabh Paliwal <[email protected]> wrote:
>
>> And now I get what you meant when you said "palindrome". You should have
>> explained that if that was "not exact palindrome"
>>
>> So yes, your solution seems correct but that is the brute force approach.
>> It runs in O(n*n*n) as there are n*n substrings and every check is linear.
>> DP solution helps save time by memoization.
>>
>>
>> On Thu, Jun 5, 2014 at 11:37 PM, Saurabh Paliwal <
>> [email protected]> wrote:
>>
>>> Ok! So I guess now we are talkng my solution. What i do is maintain two
>>> pointers i and j, i is the end of the first string and j is the beginning
>>> of the second. If both the character match, I calculate the answer for
>>> pointers i-1,j+1 and add 1 to the answer for i,j. If they don't match, I
>>> simply mark 0 as the answer for i,j. So my table if filled top-down like
>>> this. Finally, I return the maximum entry in the table.
>>> Need I explain further? If so, feel free. Should I explain what those
>>> methods do?
>>>
>>
>>
>>
>> --
>>  -    Saurabh Paliwal
>>
>>        B-Tech. Comp. Science and Engg.
>>
>>        IIT ROORKEE
>>
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-- 
 -    Saurabh Paliwal

       B-Tech. Comp. Science and Engg.

       IIT ROORKEE

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