Hi Jagdish,
I think your solution looks good. The only thing about extra space to store
count can be dealt with like this in my view:
1) Do a scan of numbers and determine the max number.
2) Create a max heap (Root is the largest number so by step 1 we have
determined the Root for the heap).
Use trie data structure ,construct it from given matrix
On Thu, May 20, 2010 at 7:23 AM, Mario Ynocente Castro ycma...@gmail.comwrote:
I don't think 1014 needs any special algorithm, if we've got an H x W
matrix, then we've got (4H+4W-2) strings in which you must look, and you can
do this
@Kaushik,
I think you can do using inorder successor also, using this successor you
can think of BST as Sorted List.
What do you say ?
Thanks,
Sathaiah
On Wed, May 19, 2010 at 11:11 AM, kaushik sur kaushik@gmail.com wrote:
@Dhilip
Is it tested ? I doubt your code won't work ?
@Rohit
@Marcio, I get your algo now. So a substring match is also a match. I get
your approach. Thank you.
Any ideas for the second problem?
On 20 May 2010 10:45, vignesh radhakrishnan rvignesh1...@gmail.com wrote:
@Mario Your estimate of no. of strings, I guess doesn't consider strings of
length
I think you have to look at this book Algorithms on Strings, Trees and
Sequences: Computer Science and Computational Biology
By Dan gusfield. It has wonderful data structure which works really fast for
string operations.
On Wed, May 19, 2010 at 4:16 PM, vignesh radhakrishnan
@Mario Your estimate of no. of strings, I guess doesn't consider strings of
length less than length H or W.
it would order(4H^2+4W^2) approximately.
I guess I 've understood it right. correct me if I'm wrong
On 20 May 2010 07:23, Mario Ynocente Castro ycma...@gmail.com wrote:
I don't think
@shachin,when u add value of root in the child of root then for insertion of
new element when heap is called(so heapify is called) so now this child
becomes root which is not desired..
plzz correct me if i m wrong
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For
Limited to trees where node key values are unique. Seems to have the
advantage of better access locality. But I suspect the average time
per node is typically greater than for Morris Traversal.
typedef struct Node_
{
int val;
struct Node_ *left, *right;
}
Node;
extern void
I was thinking about a substring matching, but if that's not what you need,
then you still have the same number of strings, but finding a matching in a
string would take quadratic time on the length of the string, instead of
linear time.
2010/5/20 vignesh radhakrishnan rvignesh1...@gmail.com
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