be list of sentences ranked in such a way that sentence
with FIRST rank is the most similar sentence in all 5 documents, then 2nd
then 3rd...
Thanks in advance.
Regards,
Anantha Krishnan
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To post
http://www.eazyupload.net/ukhyxosa
On Wed, Aug 31, 2011 at 7:29 PM, Swathi chukka.swa...@gmail.com wrote:
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Regards,
Anantha Krishnan
On Wed, Aug 24, 2011 at 11:10 AM, Dheeraj Sharma
dheerajsharma1...@gmail.com wrote:
let the nodes are stored in array like
arr[1]
arr[2]
arr[3]
.
.
.
arr[7].
where arr is a structure having int x,int y.
i=1
initally we set the root(i.e arr[i]) x and y = n/2,log
int getDistance(Tnode *root,Tnode *node,int dist)
{
if(root==NULL)
return 0;
if(root==node)
return dist;
return
getDistance(root-left,node,dist+1)|getDistance(root-right,node,dist+1);
}
Thanks Regards,
Anantha Krishnan
On Thu, Aug 25, 2011 at 2:30 AM, rohit raman.u
Hi Sarvesh,
For this problem this is my solution.
A=[1,5,6]
n=len(A)
Ans=[0,1,1,2,1,2,2,3]
count=0
for i in range(0,n):
count=count+Ans[A[i]]
print count
Is there any flaw?
Thanks Regards,
Anantha Krishnan
On Sun, Aug 21, 2011 at 4:25 PM, sarvesh saran
aquarian.thun...@gmail.comwrote
minimum element.
7.repeat these steps till end of array is reached for atleast one array.
Please let me know if you find some difficulties with my explanation.
Thanks Regards,
Anantha Krishnan
On Thu, Aug 18, 2011 at 10:42 AM, MAC macatad...@gmail.com wrote:
any suggestion on how to approach
If we move the maximum then difference will get larger but our aim is to
minimize the difference.
Thanks Regards,
Anantha Krishnan
On Thu, Aug 18, 2011 at 1:01 PM, MAC macatad...@gmail.com wrote:
as per my understanding , you are increasing the minimum value so that it
reaches closer
Yes with my approach answer will be 0.
Please check here http://ideone.com/Q4ivj.
Thanks Regards,
Anantha Krishnan
On Thu, Aug 18, 2011 at 1:33 PM, MAC macatad...@gmail.com wrote:
in the example below , answer shd be 0 , . by your apraoch this is not
commig
10,25,35
10,25,30
5
We can use hash to do all the operations in O(1) time.
Thanks Regards,
Anantha Krishnan
On Thu, Aug 18, 2011 at 5:30 PM, Ankur Garg ankurga...@gmail.com wrote:
Define a data structure , using extra memory/space , such that :
Insert(int a)
Delete(int a)
isPresent(int a)
get(int a)
All
Refer here http://ideone.com/X77wm.
On Thu, Aug 18, 2011 at 5:36 PM, Ankur Garg ankurga...@gmail.com wrote:
Can u provide a bit detail bro !!
On Thu, Aug 18, 2011 at 8:04 AM, sagar pareek sagarpar...@gmail.comwrote:
Hashing
:)
On Thu, Aug 18, 2011 at 5:30 PM, Ankur Garg
We must use
BIThttp://www.topcoder.com/tc?module=Staticd1=tutorialsd2=binaryIndexedTreesto
solve this problem with O(nlogn).
Here http://ideone.com/IaU3F is my implementation.
Thanks Regards
Anantha Krishnan
On Tue, Jul 26, 2011 at 7:18 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
You
)
{
_printTree(root+File.separator+file, depth+1);
}
return;
}
return;
}
public static void printTree(String root)
{
_printTree(root, 0);
}*
Thanks Regards
Anantha Krishnan
On Wed, Jul 27, 2011 at 1:21 PM, geek
@SkRiPt KiDdIe
I got your logic.
Nice.
Thanks.
Regards
Anantha Krishnan
On Wed, Jul 20, 2011 at 2:04 PM, SkRiPt KiDdIe anuragmsi...@gmail.comwrote:
While you are on a state do not change ur state on encountering a specific
character if u have already done so earlier.This check in addition
Can somebody please help me to print all non-redundant permutations of the
string. For ex. If string is abab the permutations are {baab, abba, abab,
baba, bbaa, aabb}
Thanks Regards
Anantha Krishnan
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For,
[ n*n!+(n+1)*(n+1)!+.+(n+m)*(n+m)! ] / (n+m+1)! = 1 - n!/(n+m+1)!
On Sun, Jul 10, 2011 at 1:00 PM, Puneet Ginoria punnu.gino...@gmail.comwrote:
n*n! = (n+1)! - n!
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,
if it is negative, find the min value, assume it is X, add by (-X)+1))
Now assume numbers are M, compute the product of the numbers and compute
M!
and check if they are equal.
does it work ?
Thanks,
Sathaiah
On Wed, Jul 6, 2011 at 11:45 AM, Anantha Krishnan
I guess it can be done in O(mlog m) TC with* heap sort*
On Fri, Jul 8, 2011 at 1:56 AM, Dumanshu duman...@gmail.com wrote:
given two sorted arrays a[m] b[2*m], each contains m elements only.
You
need to merge those two arrays into second array b[2*m].
anything better than O(m^2)
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Check this
*int isconsecutive(int a[], int n) {*
*if (n 1) {*
*return 0;*
*}*
*int max = a[0], min = a[0];*
*int i = 0;*
*
*
*int *hash = (int*) calloc(n, sizeof (int));*
*
*
*//find min and max from the array*
*for (i = 1; i n; i++) {*
*if (a[i]
lengths.
Thanks Regards
Anantha Krishnan
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Hi All,
I want to know which data structure will be efficient for a desktop search
tool similar to Ava Find http://www2.think-less-do-more.com/avafind/.
Thanks Regards
Anantha Krishnan
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http://ideone.com/qYKMW.
Thanks Regards
Anantha Krishnan
On Tue, Jul 5, 2011 at 7:52 AM, Ashish Goel ashg...@gmail.com wrote:
convert BST into DLL
refer stanford tree recursion problem
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Mon
y=*x++;
this line executes like this:
y=*x;
x=x+1; // post increment
Regards
Anantha Krishnan
On Mon, Jul 4, 2011 at 10:36 AM, amit kumar amitthecoo...@gmail.com wrote:
i think d answer sud be 10.
but it comes out to be 5.
xplain plz
On Mon, Jul 4, 2011 at 10:30 AM, Sandeep Jain sandeep6
Given a string containing 0's and 1's. One would need to find the longest
sub-string such that the count of 0's is equal to the count of 1's in the
largest sub string.
Regards
Anantha Krishnan
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Thanks.
Can you plz explain for input 1 0 1 1 0 1 1 1.
Also I want solution in O(n) TC and O(1) SC.
Regards
Anantha Krishnan
On Fri, Jul 1, 2011 at 4:13 PM, sunny agrawal sunny816.i...@gmail.comwrote:
Take an array of size of the length of the string.
fill the array positions with one where
, Jul 1, 2011 at 4:21 PM, Anantha Krishnan
ananthakrishnan@gmail.com wrote:
Thanks.
Can you plz explain for input 1 0 1 1 0 1 1 1.
Also I want solution in O(n) TC and O(1) SC.
Regards
Anantha Krishnan
On Fri, Jul 1, 2011 at 4:13 PM, sunny agrawal sunny816.i...@gmail.comwrote:
Take
@Sunny
Thanks for your algorithm :)
It works well.
http://ideone.com/MHOtR
Thanks Regards
Anantha Krishnan
On Fri, Jul 1, 2011 at 4:38 PM, sunny agrawal sunny816.i...@gmail.comwrote:
String = 1 0 1 1 0 1 1 1.
1. make the array = 1 -1 1 1 -1 1 1 1
2. after second operation
array = 1 0 1
-data;*
*} while (root-right == node || root-right == NULL);*
*if (root-right != NULL)*
*root = root-right;*
*}*
*}*
*return NULL;*
*}*
Thanks Regards
Anantha Krishnan
On Thu, Jun 30, 2011 at 2:12 PM, vikas mehta...@gmail.com wrote:
for 1 i
Check this
http://ideone.com/1I40z
On Wed, Jun 29, 2011 at 8:04 PM, Nishant Mittal
mittal.nishan...@gmail.comwrote:
segregate even and odd nodes in a singly linked list.Order of even and
odd numbers must be same...
e.g:-
i/p list is 4-1-3-6-12-8-7-NULL
o/p list 4-6-12-8-1-3-7-NULL
--
@sagar pareek
If there are more fields in the node like:
struct node{
int data;
float mark;
char ch;
struct node *link;
};
Here swapping *data* alone will corrupt the list right!!
On Thu, Jun 30, 2011 at 4:38 PM, sagar pareek sagarpar...@gmail.com wrote:
Here is one
Check this http://ideone.com/ARsJ1
On Thu, Jun 30, 2011 at 7:48 PM, oppilas . jatka.oppimi...@gmail.comwrote:
I am unable to find a test case where this particular approach fails( I
hardly thinks it's correct but anyway here it is).
We make the last element the root of the BST.
And keep
How we will get phone number of a particular person?
Thanks Regards,
Anantha Krishnan
On Wed, Jun 29, 2011 at 6:22 PM, sudheer kumar
chigullapallysudh...@gmail.com wrote:
USE TRIE
On Wed, Jun 29, 2011 at 6:10 PM, shady sinv...@gmail.com wrote:
go through the archives you will definitely
Check this
http://ideone.com/o8gF2
On Wed, Jun 29, 2011 at 10:28 PM, Swathi chukka.swa...@gmail.com wrote:
This should be very simple... follow inorder..
Inorder(Node* node, int counter, int N)
{
if(node == null)return;
Inorder(node-left, counter, N);
counter++;
if(counter == N)
{
I wish to say that we should not use any inbuilt functions.
On Wed, Jun 29, 2011 at 11:43 PM, oppilas . jatka.oppimi...@gmail.comwrote:
What I wanted to say that, it's a trivial question for algorithmic point of
view.
You could have just implemented a normal function without worrying about
http://ideone.com/oEfLE
On Thu, Jun 30, 2011 at 12:04 AM, Anantha Krishnan
ananthakrishnan@gmail.com wrote:
I wish to say that we should not use any inbuilt functions.
On Wed, Jun 29, 2011 at 11:43 PM, oppilas . jatka.oppimi...@gmail.comwrote:
What I wanted to say that, it's a trivial
(1024);*
Thanks Regards,
Anantha Krishnan
On Sat, Jun 25, 2011 at 2:16 PM, Adarsh s.adars...@gmail.com wrote:
char array[] = hello;
char *pointer = hello;
array is an array, enough to store sequence of characters and '\0'
array will always refer to same storage.
Here, pointer is initialized
Good.
On Sat, Jun 25, 2011 at 4:47 PM, RITESH SRIVASTAV
riteshkumar...@gmail.comwrote:
sizeof returns size_t values and size_t is typedef unsigned int
size_t;
but when you compare it with -1(int) ,d=-1 is converted to unsigned
int
which becomes very large (INT_MAX)
and d (INT_MAX) 7 so
Check this http://ideone.com/C8fQC
http://ideone.com/C8fQCThanks Regards,
Anantha Krishnan
On Fri, Jun 24, 2011 at 10:18 PM, Decipher ankurseth...@gmail.com wrote:
Can anybody please explain how to solve this question with logarithmic
time complexity ?
Write the code/algorithm to find
for understanding)
B[-1]=-INF A[0]=20 B[0]=10 [not in asc order]
A[-1]=-INF B[0]=10 A[0]=20 [in asc order]
We got the Kth(4th) smallest element which is 10.
Thanks Regards,
Anantha Krishnan
On Fri, Jun 24, 2011 at 11:20 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
@Anantha
can you explain
Hi All,
Can someone explain about the time complexity of Merge sort(Linked list with
billions of node)?
There is no way to find the middle of sub-list without traversing
completely.
Please clear my doubts.
Thanks Regards,
Anantha Krishnan
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,
Anantha Krishnan
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I modified Sunny's code to get Node X and Node Y.
http://ideone.com/YF9qi
Can we do better than this?
Thanks Regards,
Anantha Krishnan
On Wed, Jun 22, 2011 at 11:11 AM, oppilas . jatka.oppimi...@gmail.comwrote:
Sunny,
Can but can we modify this code to get the *node X and node Y
Hi All,
I have written code for finding diameter of a binary tree here
http://ideone.com/WHg9t
Is it correct? Do I need to make any changes there?
Thanks Regards
Anantha Krishnan
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To post
= new Node(8, NULL, g);*
*Node* a = new Node(1, b, c);*
*Root = a;*
*}*
*
*
and check here http://ideone.com/YF9qi.
Let me know the result.
Thanks Regards
Anantha Krishnan
On Thu, Jun 23, 2011 at 2:06 AM, Jitendra singh jsinghrath...@gmail.comwrote:
I think this solution
Regards
Anantha Krishnan
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Thanks.
I expect more details in implementation point of view.
Thanks Regards,
Anantha Krishnan
On Tue, Jun 21, 2011 at 6:41 PM, sunny agrawal sunny816.i...@gmail.comwrote:
@Piyush
good to start with
But i think a recursive O(n) is possible
in downward calls pass sum from root to node
@sunny agrawal
*Thanks a lot.*
*Great code. I got the logic now.Thanks again.*
*
*
Thanks Regards,
Anantha Krishnan
On Tue, Jun 21, 2011 at 11:52 PM, sunny agrawal sunny816.i...@gmail.comwrote:
see this
*
https://ideone.com/1ZtIq*
On Tue, Jun 21, 2011 at 10:23 PM, Anantha Krishnan
));
}
Thanks Regards,
Anantha Krishnan
On Thu, Jun 9, 2011 at 4:54 PM, Navneet Gupta navneetn...@gmail.com wrote:
Ohh. Missed out the nlogn condition you mentioned. It will do but in n^2
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