Given that the list is in sorted order. Let us assume that the list in
the form of an array A[1...n].
Case 1: If n is odd. Then the median is A[(n+1)/2]. Set MEDIAN:=A[(n
+1)/2.
Case 2: If n is even. Then the median is (A[n/2]+ A[n/2 +1])/2. Set
MEDIAN:=(A[n/2]+ A[n/2 +1])/2.
Assuming that the
/** find the Middle element in the Array without sorting it.
* This function uses a modified version of QuickSort, where we
* only consider the one half of the array.
* This is a recursive function where we look at some section of the array
* (Concerned Array) at a time.
*
* @param low
Please ignore my previus mail as there was some typo mistake.
/** find the Middle element in the Array without sorting it.
* This function uses a modified version of QuickSort, where we
* only consider the one half of the array.
* This is a recursive function where we look at some section of
Hi,
I'm a co-editor of a special issue of the Journal of Genetic
Programming and Evolvable Machines which looks back at past progress
and future possibilities in this amazing area of computer science and
machine intelligence. Among lots of goodies it includes a review of
human competitive results
Hi Anand,
Can you please explain your code? What is the magic
number 10 in
if(k == 10)
{
printf(String Matched\n);
}
in your code?
What does while loop do in your code? Can you please write a comment?
-Thanks in advance,
Bujji
#include stdio.h
#include
A company has many employees each employee is led by only 1 person
except for the CEO (who has no boss). The cost to the company of an
employee is the sum of his salary plus the cost to the company of all
the people led by him. Given is the following structure :
Employee
i donot think that the link provided is for the same problem, the link
provides a solution to balance a tree whereas this problem is to merge
two BST without any limitation on the balance factor
having said that th balancing the tree itself is an interesting
problem and i must say that the
I have a TREE class like this.
class tree
{
public:
struct node *root;
struct node *head;
tree()
{
root=NULL;
head=NULL;
}
void insert(int num);
void inorder(node *temp);
int depth(node *temp);
if it is a simple BT then you can simply attach the root to either
child ( which is null ) of other tree . just simply go leftmost and
then assign root of other tree as left child, as suggested by Gene.
On Jul 27, 8:23 am, Gene gene.ress...@gmail.com wrote:
You actually only need a singly linked
It is just an Implementation of KMP string match Algorithm.
In the first section, I am find the prefix function π for a pattern which
encapsulates knowledge about how the pattern matches
against shifts of itself.
This information can be used in second section to avoid testing useless
shifts for
Think hard! It's for a slightly different problem, but the same
algorithm works fine. It converts an arbitrarily skewed tree to a
perfectly balanced tree. So you can just walk the two input trees,
deconstruct them into sorted lists (a fully skewed tree is just a
list). Merge the two lists into a
I think you have confused the statement of this problem. The (in
sorted order) comment makes no sense because a median is exactly one
number. One number is always sorted.
After every stream element is read, you want to be able to get the
median of all elements read so far.
You're correct that
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