@utsav u haven't initialized match anywhere so ur algo fails
On Thu, Jun 7, 2012 at 2:50 AM, utsav sharma <utsav.sharm...@gmail.com>wrote:
> i think it can be solved using DP
> word="bcdf" take hash of word h[b]=1 h[c]=2 h[d]=3 h[f]=4
> given 2d matrix m[][]=
> {b c b e f g h
> b c d f p o u
> d f d f g k p }
>
> take another matrix match[][]
> if( h[ m[i][j] ] > 0 ) //if this char is in word then
> {a=h[ m[i][j] ];
> if (match[i-1][j] ==a-1 || match[][]=a-1 || match[][]=a-1 ) check
> prev element of row / diagonal /column
> match[i][j]=a;
> }
> else if char is not matched, then match[i][j] will contain longest prefix
> match(as in KMP).
> if at any instance we get match[i][j]==no. of chars in word then we will
> backtrack it to get the string.
> correct me if i'm wrong !!
>
> On Wed, Jun 6, 2012 at 10:39 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> i did this question long time back....
>> well simple brute force check can be done....you can keep one flag
>> matrix of same size to avoid necessary recursion.
>>
>>
>> On 6/6/12, Ashish Goel <ashg...@gmail.com> wrote:
>> > WAP to find a word in a 2D array. The word can be formed on
>> > row/col/diagnal/reverse diagnal
>> >
>> > Best Regards
>> > Ashish Goel
>> > "Think positive and find fuel in failure"
>> > +919985813081
>> > +919966006652
>> >
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>
>
> --
> Utsav Sharma,
> NIT Allahabad
>
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regards,
Bhaskar Kushwaha
Student
CSE
Third year
M.N.N.I.T. Allahabad
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