hi, i need to find the number of nodes at each level of a binary tree..the
binary tree may not be balanced..
output:
Level 0 - 1 node
Level 1 - 2 nodes
Level 2- 3 nodes
and so on..based on the tree structure..I am not able to count at each
level..pls suggest a way to do that.
--
Harshal
do BFS.
On Oct 23, 6:31 pm, Harshal hc4...@gmail.com wrote:
hi, i need to find the number of nodes at each level of a binary tree..the
binary tree may not be balanced..
output:
Level 0 - 1 node
Level 1 - 2 nodes
Level 2- 3 nodes
and so on..based on the tree structure..I am not able to
use 2 datstructure : TRIE and an array of words along with their
frequency.
we can solve it in following step:
1) scan each word of the large file. insert the current word into the
TRIE if not present already. update the frequency.
using TRIE, the time complexity is O(L), where l is the
When visiting appropriate vertex v, increment counter +
+levels[current_depth] and go further.
You may done this using DFS or BFS.
On 23 окт, 17:31, Harshal hc4...@gmail.com wrote:
hi, i need to find the number of nodes at each level of a binary tree..the
binary tree may not be balanced..
@Praveenit is not possible..in a BST *all the nodes* on the right
subtree are greater than the node :)
On Sat, Oct 16, 2010 at 3:26 PM, Praveen Baskar praveen200...@gmail.comwrote:
@nishaanth: wat if the left child of the right node has a negative value
On Thu, Oct 14, 2010 at 11:12 AM,
It is nothing but a common subsequence problem...isnt it ?
On Wed, Oct 13, 2010 at 3:42 PM, Mridul Malpani malpanimri...@gmail.comwrote:
@ ankit agarwal, you are right. thanx man.
On Oct 13, 11:37 am, prodigy 1abhishekshu...@gmail.com wrote:
Let I,Q be input array,query array respectively.
I've written a kart-trie in php. You can easily extend yourself the
payload to count the word frequency.
http://sourceforge.net/projects/kart-trie
After you build your dictionary from your large file, you can easily
find the highest frequency be recursively search the trie. It should
be faster
http://www.series.el3lam.com/293.html
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Parallel edges mean?
On Thu, Oct 21, 2010 at 3:22 AM, Algorithimist yogi15...@gmail.com wrote:
Design an algorithm to determine whether a graph has any parallel
edges in time proportional to E + V.
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The biggest weakness of the trie is the amount of space it wastes -
chances are there will be runs of characters with a word only at the
end, meaning a bunch of extra levels of the tree that aren't needed.
The PATRICIA trie, or radix trie, attempts to address this by allowing
the 'decision' at
Given n point on the plane, find out whether any 3point on the same line.
How to use recursion to solve the problem? Could you help me find the
algorithm and give the time complexity?
Bests,
Claire
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Is there any specific need to use recursion?
One alternate is to find slope and constant (m and c) for every pair of
points and same value of m c will specify the points on the same line.
Time complexity is O(n*n).
On Sat, Oct 23, 2010 at 4:31 PM, Meng Yan mengyan.fu...@gmail.com wrote:
there are (n*(n-1))/2pairs of points. I think if we use your method, the
time complexity should be O(n^4).
Is it possible to put all points into k different domain and using
T(n)=T(n/k)+f(n) to solve this problem?
On Sat, Oct 23, 2010 at 7:51 PM, preetika tyagi preetikaty...@gmail.comwrote:
Is
I gave an O(n^2 log n) algorithm to find the maximal number of
collinear points in a set is given in
http://groups.google.com/group/algogeeks/msg/d329dda12b332dd1.
A fairly simple modification could answer the question as to whether
any three points are collinear.
Dave
On Oct 23, 6:31 pm, Meng
Thank you!
By the way, to do the 'sort s(i+1:n)', if I use counting sort, I think it
should be better.
imax = 0 // location of longest string of duplicate slopes
lmax = 0 // length of longest string of duplicate slopes
smax =
You have to scan every pair of points only once to get the value of 'm' and
'a', so the time complexity would be O(n^2).
On Sat, Oct 23, 2010 at 6:22 PM, Meng Yan mengyan.fu...@gmail.com wrote:
there are (n*(n-1))/2pairs of points. I think if we use your method, the
time complexity should be
i think it is possible nishaanth
please do take a look at this example
-10
/\
-11 8
/\
-5 10
-5 is greater than -10
On Sat, Oct 23, 2010 at 11:19 PM, nishaanth
@Praveen- In this case, we will not ignore the right subtree of the root
(-10, which is less than zero) while traversing the tree.
On Sat, Oct 23, 2010 at 9:06 PM, Praveen Baskar praveen200...@gmail.comwrote:
i think it is possible nishaanth
please do take a look at this example
@Meng Yan: s is an array of real numbers, not integers, so perhaps a
counting sort is not applicable.
Dave
On Oct 23, 10:51 pm, Meng Yan mengyan.fu...@gmail.com wrote:
Thank you!
By the way, to do the 'sort s(i+1:n)', if I use counting sort, I think it
should be better.
@Preetika: Then you have to look for duplicates in an array of n(n-1)/
2 real numbers. I think this takes the complexity above O(n^2).
Dave
On Oct 23, 10:54 pm, preetika tyagi preetikaty...@gmail.com wrote:
You have to scan every pair of points only once to get the value of 'm' and
'a', so the
Do level order traversal using two queues.
On Oct 23, 8:19 pm, juver++ avpostni...@gmail.com wrote:
When visiting appropriate vertex v, increment counter +
+levels[current_depth] and go further.
You may done this using DFS or BFS.
On 23 окт, 17:31, Harshal hc4...@gmail.com wrote:
Can be done in O(n^2) time using the slope as people suggested above.
1- Sort the points in increasing order of x cord. O(nlogn)
2- prepare a n*n matrix A where A[i,j] = slope( point(i), point(j) ) -
O(n^2) [Note that point i and j are sorted in increasing order of x]
3- find a pair of A[i,j]
For duplicates, either we can sort or use a Hash Set at the cost of extra
space. While traversing all the points for slope etc calculation, insert the
point into hash set. If a point already exists, skip it. Like this, we are
calculating the slope and keeping track of distinct points in a single
@Ankit- I think it can be done using a single queue also.
2010/10/23 ankit agarwal ankit.agarwal.n...@gmail.com
Do level order traversal using two queues.
On Oct 23, 8:19 pm, juver++ avpostni...@gmail.com wrote:
When visiting appropriate vertex v, increment counter +
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