there is at all any difference between a Singleton class and one with
all static member (ie methods and attributes )
We could not find any instance where
'all static member class' would not achieve
the same functionality as class
properely implementing Singleton pattern ??
For eg
use a dictionary (available in C#... basically a collection of generic
key-value pairs where the key lookup is O(1) - hashed internally)
since number that occurred first should be listed first when
frequencies are the same, u need to record the first occurrence of
each number as well. Hence, the
Problem is to find print all the nodes at the same levels of a binary
tree.Inputs given are root pointer and the pointer to the node (let it
be A)whose level elements need to be printed.
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A rooted binary tree with keys in its nodes has the binary search tree
property (BST property) if, for every node, the keys in its left
subtree are smaller than its own key, and the keys in its right
subtree are larger than its own key. It has the heap property if, for
every node, the keys of its
Lets assume that the tree node has two keys K1 and K2.
K1 satisfies the BST property
K2 satisfies the Max Heap Property.
Our problem is to build a binary tree which satisfies both the
properties.
For a maximal heap the root node must be the maximum.
So we find the node which has the K2 max. And
Hi,
Very good question.
i think very knows brute-force soln of n^3 time complexity.
here is a alternative method/soln of O(n^2 * log(n)) time complexity Algo
1) pick any number from list A (One by one).. let say 'x'
2) pick any number from list B (one by one ).. let say 'y'
3) z=x+y
no not sorted
On 15 December 2010 11:37, Anand anandut2...@gmail.com wrote:
@Divya, I think you ask this question before. Not sure of the answer though
:)
On Tue, Dec 14, 2010 at 9:06 PM, Soumya Prasad Ukil ukil.sou...@gmail.com
wrote:
Are the linked list sorted?
On 14 December 2010
Given an array of size n wherein elements keep on increasing
monotically upto a certain location
after which they keep on decreasing monotically, then again keep on
increasing, then decreasing
again and so on. Sort the array in place (ie. using only O(1) extra
memory).
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Just the opposite.
All the operands are evaluated from left to right, and the right most
operand is returned as the value of whole comma expression.
But remember the operator precedence. Comma is lower than assignment.
So c=--a, b++ - c is equivalent to c=--a; b++ -c;
While c=(--a, b++ -c) is
wouldn't your algo take n^3 time as well given the fact that the lists are
not sorted? searching for the matching value fr z should be taking O(n) time
and to do so for each pair of x and y shall take O(n^2).
Please correct me if i got something wrong here
On Wed, Dec 15, 2010 at 11:44 AM,
@prashant
How does log(n) come in your algo? I think yours is O(n^3).
Using Hashtable,if allowed, reduces complexity.
On 14 December 2010 22:14, Prashant Kulkarni prashant.r.k...@gmail.comwrote:
Hi,
Very good question.
i think very knows brute-force soln of n^3 time complexity.
here is a
First determine the point from which array becomes decreasing. Then In-place
merge sort.
On 14 December 2010 23:22, divya sweetdivya@gmail.com wrote:
Given an array of size n wherein elements keep on increasing
monotically upto a certain location
after which they keep on decreasing
@ankur,saurabh,soumya..
ya ankur i forget to remove range from dare also no need to find
range for this..\
put size-1 intead of range so that malloc will alocate the memory to
all elements in array ..no hope its fine...
what i did is that
i took counter array thta cvontains the no of
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Subject: [Lovers India] @~|| DFFERENCES BETWEEN LOVE AND LIKE ||~@
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we_love_in...@googlegroups.com,
HI HOW ARE YOU
On Tue, Dec 14, 2010 at 7:45 PM, divya sweetdivya@gmail.com wrote:
an array contain +ve and -ve element, find subarray whose sum =0;
Lets take input array as a[]={-3,2,4,-6,-8,10,11}
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Algorithm
refer topcoder tutorials..dp
On Wed, Dec 15, 2010 at 12:12 PM, Ankur Khurana ankur.kkhur...@gmail.comwrote:
how do we find the complexity of DP program ? i know it cn be done
using DP states , but the gien complexity is n^2 . i am not sure how
to compare that
On Tue, Dec 14, 2010 at
It's a subset-sum problem, I guess.
On 15 December 2010 04:12, parth panchal parthpancha...@gmail.com wrote:
HI HOW ARE YOU
On Tue, Dec 14, 2010 at 7:45 PM, divya sweetdivya@gmail.com wrote:
an array contain +ve and -ve element, find subarray whose sum =0;
Lets take input array as
Have a look : http://geeksforgeeks.org/?p=1488
On 15 December 2010 05:19, Saurabh Koar saurabhkoar...@gmail.com wrote:
@ Bittu:
Lets analyze your code with iterations:
the array contains 1 3 3 1 2 3 5 2 3
count contains 0 2 2 4 0 1 0 0 0
now 1st iteration:
i=8,7,6 the inner loop
A sequence of numbers is called a zig-zag sequence if the differences
between successive numbers strictly alternate between positive and
negative. The first difference (if one exists) may be either positive
or negative. A sequence with fewer than two elements is trivially a
zig-zag sequence.
For
we can use binary search algorithm to find the z
( sorting ll take O(n log n ) but it is very less compared to n^2 * log n )
-- Prashant Kulkarni
On Wed, Dec 15, 2010 at 3:32 PM, Ankur Murarka
ankur.murarka@gmail.comwrote:
wouldn't your algo take n^3 time as well given the fact that
@ankur: the operation for each cell in dp array is O(1) so since we fill n^2
cells the total complexity is O(n^2)
On Wed, Dec 15, 2010 at 10:12 AM, Ankur Khurana ankur.kkhur...@gmail.comwrote:
how do we find the complexity of DP program ? i know it cn be done
using DP states , but the gien
Here is an O(N^2) algorithm:
1. Sort both A and B. (O(NlogN) or O(N^2), either will work)
2. For each element C[k] in C, find if there is such (i,j) pair that
A[i] + B[j] = -C[k]:
Start with i = 0 and j = N-1, and loop through the following:
a. if A[i] + B[j] -C[k], then j = j-1;
main()
{
int a[5]={1,3,6,7,0};
int *b;
b=a[2];
coutb[-1];
}
the value of b[-1] is
a.1
b.3
c.-6
d.none
give me the answer and explain why ??
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The answr is 3. Arrays can have negative elements. The address is the
address of base - address of int.
On Wed, Dec 15, 2010 at 10:56 PM, siva viknesh sivavikne...@gmail.comwrote:
main()
{
int a[5]={1,3,6,7,0};
int *b;
b=a[2];
coutb[-1];
}
the value of b[-1] is
a.1
b.3
c.-6
Is it that the first case its a pointer to a pointer. So size of pointer to
a pointer is 4.
In next case , Its the size of array. So its 40. Is the explanation correct?
On Wed, Dec 15, 2010 at 9:43 AM, rahul rahulr...@gmail.com wrote:
you would like to read peter ven der linden.(deep C
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