how about google sparsemap and densemap ; they are good with larger set of
data . used it with iris and imdb dataset for clustering.
On Tue, Jul 6, 2010 at 4:46 PM, Jitendra Kushwaha
jitendra.th...@gmail.comwrote:
What about STL map
I have used it. but for small data..
It is a standard one
use array element as index in linear traversal and der index pointing value
as negative when u wll get pointing value negative mean that array index
position is 1st repeated element.
On Wed, Jul 7, 2010 at 10:18 AM, Anil C R cr.a...@gmail.com wrote:
do you need an algorithm which is O(1) in
@harit why pipes are fastest...plzz explain a bit
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I think you can use counting array in this algo. And keep checking
whenever the count of any element goes ahead of 1, it will be the
first repeating element.
Is it something more difficult than this ?? Please correct me if
anything is wrong in my thought.
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I don't understand how to constuct the suffix tree in less than O(n^2) can
anyone explain me this?
On Tue, Jul 6, 2010 at 10:03 AM, Jitendra Kushwaha jitendra.th...@gmail.com
wrote:
@Ankit Narang
Think about your algo it is not a O(n). First of all you wont get
solution comparing start of
m!=n
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if space complexity is O(1) then hash map will not work..
On Wed, Jul 7, 2010 at 10:54 AM, Ashish Goel ashg...@gmail.com wrote:
have a hash map trace through all the elements to store the count
now trace through the array again and return the element whose count is
found to be 2 as the first
Firstly it is srm 475, the following link has the problem
http://www.topcoder.com/stat?c=problem_statementpm=10878rd=14156
@crazysaikat : Sorry for misconstruing you. As this group is public it is
better not to post problems of a srm while it is running.
Apart from discussing it here, if you need
@ashish cant u make use of char *p[30] for 2 d array
On Wed, Jul 7, 2010 at 9:34 AM, Ashish Goel ashg...@gmail.com wrote:
char name[][10] is a auto variable on stack, so no pointers here
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
Hi Gaurav,
yup solution is quite right but that is the worst casebut if wanna that
solution in O(n) time complexity and O(1) space complexity but in ur
solution it is O(k) space complexity.
On Wed, Jul 7, 2010 at 7:17 AM, Gaurav Singh gogi.no...@gmail.com wrote:
I think you can use
this the same hamming distance problem...this can be done thorugh
trie.pls check archives this has been discussed before..
On Wed, Jul 7, 2010 at 10:12 AM, Ashish Goel ashg...@gmail.com wrote:
use levenstein distance algo
http://en.wikipedia.org/wiki/Levenshtein_distance
Best
ya i want inplace soln
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you have misinterpreted the questn.. 1 simply means 1 occurences we have
to do this inplace
On Wed, Jul 7, 2010 at 10:37 AM, Ashish Goel ashg...@gmail.com wrote:
i thought the same way, but in this case 4 is being repeated twice, but is
not nullified
it is ctually getting nullified,
There are more efficient ways of doing this.
Refer wiki http://en.wikipedia.org/wiki/Cycle_detection#Brent.27s_algorithm
On Wed, Jul 7, 2010 at 9:16 AM, Dave dave_and_da...@juno.com wrote:
@Anand. You've described one way to do it, and maybe the most
efficient way, but not the only way.
Dave
@ashgoel
Thanks for the reply. Any web link or book / chapter from which I can
read more about the nearest word finding approach using trie you
mentioned?
Thanks in advance.
Sourav
On Jul 4, 11:12 am, ashgoel ashg...@gmail.com wrote:
bloom filters are used for approx matches, however, if you
good problem modeling :)
On Tue, Jul 6, 2010 at 10:21 AM, Jitendra Kushwaha jitendra.th...@gmail.com
wrote:
problem can be converted in a graph question
we have nC2 paths in n node (here n represent total points)
now do travelling salesman on this graph. And the question is done
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ur understanding of problem is fully correct
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@all
Please find my O(n) time and O(1) space implementation for this
problem.
Summary of approach: let i be start for first sorted part and j be
start of next sorted part. move i as long as a[i] is less than a[j] as
that means things are sorted.
if a[i] a[j], swap them and increment i.
@Amit
You should me able to find waether there is a loop or not in a LL with
any value of m and n. But if you have to find where the loop is, then
you have to chose m and n such that gcd(m,n)=1, as Dave said. Consier
the example below
I think pipes are fastest as the other end process will be trying to read
the from fds always(waiting for the input).
semaphores,monitors all these need certain condition to meet so that they
get ++/-- , notify.
Signal's can be masked!!.
.
Satya
On Wed, Jul 7, 2010 at 8:13 AM, sharad
not to complicate the question, if it is sorted, then its simple!!
.
Satya
On Wed, Jul 7, 2010 at 3:05 PM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
you have misinterpreted the questn.. 1 simply means 1 occurences we
have to do this inplace
On Wed, Jul 7, 2010 at 10:37 AM,
the approach came just from the top of my head :)
please check suffix trie @ http://www.allisons.org/ll/AlgDS/Tree/Suffix/
if you understand this as well as levensthien distance concept, you would
understand how i thought of this solution
Best Regards
Ashish Goel
Think positive and find fuel
I believe a merge sort requires O(n) space complexity. Is stack spce
counted towards space complexity? If not, I imagine you could write a
merge sort recursively, so that the explict space usage has O(1)
complexity.
On Jul 2, 2:37 pm, Abhishek Sharma jkabhishe...@gmail.com wrote:
I think its
for(i=0 to n-1)
if( binarysearch(i,n-1,1) + 1)
count++
print count.
binarysearch(first,last,item)
if(1 is there)
return mid
else
return -1.
similarly we can go for coloumns.
o(nlogn)
On 7/5/10, divya jain sweetdivya@gmail.com wrote:
i think u need to visit every element atleast
@sharad
When you say you want first repeating element, do u mean first in the
sense in which numbers are layed out in the array (i mean moving from
left to right in the array, the first element, =K, that is repeating)
or the first smallest element that is repeating? for example in the
given
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