Using Dynamic programing(Longest common subsequence logic) we can solve this
problem in O(nm) where n is the length of the first string and m is the
length of second string. Last element of matrix which the length of the
string that matches.

http://codepad.org/cyiKEMSF

<http://codepad.org/cyiKEMSF>

On Sun, Aug 1, 2010 at 8:13 AM, srikanth sg <srikanthini...@gmail.com>wrote:

> plz .. if any one knows dp solution then tell ...
>
>
> On Sun, Aug 1, 2010 at 7:31 AM, Ashim Kapoor <ashimkap...@gmail.com>wrote:
>
>> I am not sure, but can I do this using a suffix trie ? any comments ?
>>
>>
>>
>>
>> On Sun, Aug 1, 2010 at 2:29 PM, Ashish Goel <ashg...@gmail.com> wrote:
>>
>>> solution could be to find the charcter position from both sides for each
>>> char of s2
>>> then from the 2*n array, find the smallest index from left and largest
>>> from right, within these two indexes all chars would be there
>>>
>>> one case where one of the chars can be missing can be found if a row in
>>> this 2-d array remains unmodified
>>>
>>>
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>>
>>> On Sat, Jul 31, 2010 at 10:22 PM, Nikhil Jindal 
>>> <fundoon...@yahoo.co.in>wrote:
>>>
>>>> At the moment, I can only think of an O(n^3) algo.
>>>> Maybe if you can find a hash function which computes the hash value
>>>> depending on the unique characters the string contains, you can reduce it 
>>>> to
>>>> O(n^2).
>>>>
>>>>
>>>> On Sat, Jul 31, 2010 at 7:09 PM, srikanth sg 
>>>> <srikanthini...@gmail.com>wrote:
>>>>
>>>>> given two string , find the minimum  width in string 1 containing the
>>>>> all characters of string 2 , they may present in different order
>>>>> string 1-HELLO
>>>>> string 2- LE
>>>>> answer-2
>>>>>
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>>>>
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