On May 14, 9:46 pm, Yalla Sridhar sridhar2...@gmail.com wrote:
@atul linked list can be converted to array with O(n) both space @ time
overhead. then tree can be built in O(n)
This is true but is not the simplest solution.
As pointed out earlier by me and also Divya Jain
the best solution is
@sharad : Can you explain with same 'amazon' example for key mapping.
if we have O(K) hash map, how would we map keys as We need to
'remember' mapping to print things back .
e.g. ASCII valueof (C1)% sizeof(string) could be equal to
valueof(C2)%sizeof(string) where strins is C1C2...
@divya your
@Rohit
Can u pass on thje link of morris inorder
On 5/15/10, Rohit Saraf rohit.kumar.sa...@gmail.com wrote:
there is something called morris inorder traversal.
credits to donald knuth
On 5/15/10, kaushik sur kaushik@gmail.com wrote:
Hi Friends
I have encountered the question in
Thanks Rohit!
On Sat, May 15, 2010 at 7:13 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
there is something called morris inorder traversal.
credits to donald knuth
On 5/15/10, kaushik sur kaushik@gmail.com wrote:
Hi Friends
I have encountered the question in sites - Given a
@Divya I think your solution is correct. To do in constant time , we
can use BST itself instead of storing in array.
Have two array , make first point to left most , make second point to
right most member, now start your algorithm while making
these two pointers move. Left pointer should move in
@Modelling expert.
#includeiostream
#includemap
#includestring
using namespace std;
int main()
{
string s=amazon;
int i=0,j=0;
mapchar,intmp;
for(i=0;is.length();++i)
{
mp[s.at(i)]++;
}
will your code work for tree attached and for sum =40??
On Fri, May 14, 2010 at 11:44 PM, jalaj jaiswal
jalaj.jaiswa...@gmail.comwrote:
Strategy: subtract the node value from the sum when recurring down,
and check to see if the remaining sum is 0 when you run out of tree.
let sum be
@ALL,
Here order of the string is preserved, if we use hash or bst, how you
preserve the string order ?. Do you search again for each char ?
@jalaj, do we need to give importance to order or not ?
amazon = a2m1z1o1n1
Thanks,
Sathaiah
On Sun, May 16, 2010 at 5:40 PM, sharad kumar
would it be possible to come with the abstract solution with the following
idea.
1. Use the inorder traversal, (we know this it prints elements in ascending
order)
2. Use the same inorder traversal (inorder_rev) with the left, right
change,( this will print in descending order)
3. Can we use this
suppose if i give
Ssmall:es
Sbig:he's a algogeek and he's rocking
wat will be o/p?
On Sun, May 16, 2010 at 8:12 PM, divya sweetdivya@gmail.com wrote:
You r given a large string of characters lets call it Sbig. Then there
is a small set of characters lets call it Ssmall. You have to find
output ll be : e's
On 16 May 2010 20:17, sharad kumar aryansmit3...@gmail.com wrote:
suppose if i give
Ssmall:es
Sbig:he's a algogeek and he's rocking
wat will be o/p?
On Sun, May 16, 2010 at 8:12 PM, divya sweetdivya@gmail.com wrote:
You r given a large string of characters
use dp to solve this.
On Sun, May 16, 2010 at 8:17 PM, sharad kumar aryansmit3...@gmail.comwrote:
suppose if i give
Ssmall:es
Sbig:he's a algogeek and he's rocking
wat will be o/p?
On Sun, May 16, 2010 at 8:12 PM, divya sweetdivya@gmail.com wrote:
You r given a large string of
@navin naidu
like LCS in CLRS???
On Sun, May 16, 2010 at 8:20 PM, divya jain sweetdivya@gmail.comwrote:
output ll be : e's
On 16 May 2010 20:17, sharad kumar aryansmit3...@gmail.com wrote:
suppose if i give
Ssmall:es
Sbig:he's a algogeek and he's rocking
wat will be o/p?
On
@Navin: and that works ! :)
@all : i am sure no heuristic/greedy strategy can be applied.
@divya : did you check your array partitioning algorithm with my example !
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
Either the root will be included or it will not be. If it's not, then it's
equivalent to solving the problem on the subtrees.
So let's consider the case when root node is included
Now we keep track of A[node1,node2,reqdWeight]
reqdWeight is the sum of wt reqd from paths starting from node1 and
@anurag : won't work
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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Algorithm Geeks
Read it up : http://www.eecs.umich.edu/~qstout/pap/CACM86.pdf
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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You received this message because you are
@Nikhil : For that size of subtrees at all nodes needs to be maintained. And
in that case this is a trivial problem.
@Sathaiah : See my solution :)
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
@sharad : Thanx !! But is it really O(K) ?
if you use mapchar,int for say AMAZON you need 5 (chars) + ( 5*4)
( ints) = 25 bytes. + some more space due to MAP structure
One more char and you are crossing 26 bytes solution provided by
Jalaj. Think about storing Allodoxaphobia
A= 3 , L = 2 O = 3 , D
sorry , there was a typo
'Have two array read it as Have two pointers
-Manish
On May 16, 1:04 pm, Modeling Expert cs.modelingexp...@gmail.com
wrote:
@Divya I think your solution is correct. To do in constant time , we
can use BST itself instead of storing in array.
Have two array , make
@anurag: wont work, consider the following case: -99 0 2 -1 99
On Sun, May 16, 2010 at 9:12 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
@anurag : won't work
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and
@Sharad: yup
On Sun, May 16, 2010 at 8:36 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
@Navin: and that works ! :)
@all : i am sure no heuristic/greedy strategy can be applied.
@divya : did you check your array partitioning algorithm with my example !
How many ROTATIONS are required to convert any n-node BST to convert to any
other n-node BST ?
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