if ( ten's place is != 9)
Decrement the unit place by 1 and increment the ten's place by 1.
else
Increment MSD(Most significant digit) by 1 and decrement next digit by 1.
example:-
712 721
897 987
On Sun, Oct 9, 2011 at 6:51 PM, wujin chen wujinchen...@gmail.com wrote:
@Navneet，
Use K-Median theorem with k=N
On Mon, Nov 7, 2011 at 11:33 PM, Gene gene.ress...@gmail.com wrote:
Can you explain how a heap helps get the answer? Just to put the
elements in a heap requires O ( N^2 log (N) ) time.
On Nov 7, 4:12 pm, vikas vikas.rastogi2...@gmail.com wrote:
I think the
Say you define ur matrix in M
then
if (M(i,j) = 1)
Sq(i,j) = min(Sq(i-1,j),Sq(i-1,j-1),Sq(i, j-1)) + 1
else
Sq(i,j) = 0
On Tue, Nov 8, 2011 at 7:27 AM, vikas vikas.rastogi2...@gmail.com wrote:
try this:
sq(i, j)= k is maximum sqare possible ending at i, j and has
length k in the
--
Regards
Kumar Raja
M.Tech(SIT)
IIT Kharagpur,
10it60...@iitkgp.ac.in
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
Hi to find running median from a stream of random generated numbers I have
heard of the 2 heap ( min and max heap ) solution but I fail to understand
it...could someone please explain with a small example or so ??
thanks!
--
People often say that motivation doesn't last. Well, neither does
unlink each node in original tree in postorder, and insert these nodes in
new bst tree
surender
On Tue, Nov 8, 2011 at 4:48 AM, vikas vikas.rastogi2...@gmail.com wrote:
@ Above
no need to have another array or nything
binTreeToBST(node *root)
{
if(!root )return;
node *newRoot;
apti + technical + 3 prog (mainly linked list)
On Tue, Nov 8, 2011 at 3:09 PM, kumar raja rajkumar.cs...@gmail.com wrote:
--
Regards
Kumar Raja
M.Tech(SIT)
IIT Kharagpur,
10it60...@iitkgp.ac.in
--
You received this message because you are subscribed to the Google Groups
Algorithm
using a max heap and min heap we can find the median
On Tue, Nov 8, 2011 at 10:59 AM, Arun Vishwanathan
aaron.nar...@gmail.comwrote:
Hi to find running median from a stream of random generated numbers I have
heard of the 2 heap ( min and max heap ) solution but I fail to understand
read this
http://www.geeksforgeeks.org/archives/14873
On Nov 8, 10:29 am, Arun Vishwanathan aaron.nar...@gmail.com wrote:
Hi to find running median from a stream of random generated numbers I have
heard of the 2 heap ( min and max heap ) solution but I fail to understand
it...could someone
How to find the indexes of a repeated element in the sorted array in
*log n*time..
e.g.
a: 1 3 4 4 5 6 7 8 8 9 9 9 10
x=9
ouput is 10 and 12 (indexing from 1)
--
Regards
Kumar Raja
M.Tech(SIT)
IIT Kharagpur,
10it60...@iitkgp.ac.in
--
You received this message because you are
I googled around and couldn't find a good source for the K-Median Theorem.
So, could you provide a good link?
On Tue, Nov 8, 2011 at 2:55 PM, sagar sindwani sindwani.sa...@gmail.comwrote:
Use K-Median theorem with k=N
On Mon, Nov 7, 2011 at 11:33 PM, Gene gene.ress...@gmail.com wrote:
Can
Gene,
we are not creating heap here but using the sorted matrix as heap
itself.
so use same logic of removing items from heap , in this case you have
either right/bottom to reconstruct the structure. removing N^2 /2
items will give you the median.
On Nov 7, 11:03 pm, Gene gene.ress...@gmail.com
@sagar, your method is correct for three digits, but since the
question says the length could be upto 1000, the MSD in those cases
will change
Eg. 23998 24997 and not 32998
So in your algo, MSD should be replaced by while moving away from
tens, hundreds, thousands... and so on, the first
I have studied about it ,but could not understand why their construction is
in linear time and support so many operations with less time complexity. i
am quite confused. suggest me some good resource to learn about them with
proper proof of time complexity for each and every operation..
--
Using binary search, first find the first occurrence of x. Using this first
occurrence run another binary search to find the last occurrence of x.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to
Thanks rahul
On 8 November 2011 03:32, rahul sharma rahul23111...@gmail.com wrote:
apti + technical + 3 prog (mainly linked list)
On Tue, Nov 8, 2011 at 3:09 PM, kumar raja rajkumar.cs...@gmail.comwrote:
--
Regards
Kumar Raja
M.Tech(SIT)
IIT Kharagpur,
10it60...@iitkgp.ac.in
--
Check out Ukkonen's algorithm. You can build a suffix tree in O(n). Good
stuff. I would start there
On Tue, Nov 8, 2011 at 10:41 PM, kumar raja rajkumar.cs...@gmail.comwrote:
I have studied about it ,but could not understand why their construction
is in linear time and support so many
Vikas,
The cost of removing elements from the matrix is O(N) it self. So by
removing N^2/2 elements, the complexity of your algo should be N^3. However
there are well-known algo for median in O(N) time in this case O(N^2)
*because there are N^2 elements.
On Tue, Nov 8, 2011 at 6:58 AM, vikas
Suppose the array is not sorted and we have to find if an element has
occurred earlier or not; and if yes, then remove
it.what is the best achievable time and how?
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
19 matches
Mail list logo
