Very sorry the mistake. I'm not familiar with your name, and few
computer science students where I live are women.
The first detail is that binary search will only work if you assume no
duplicates. This is easy to see if you envision an input that is all
1's except for a single 2. In order to d
In both C and C++, the result of a right shift of a signed value is
implementation specific. The vacated bits can be filled either with
zeros or with copies of the sign bit. Portable code must not depend on
the implementation, but must work with either implementation choice.
Thus, the result could
in case of -ve numbers irrspectiveof size of compilerits repesented as
in remanig space apart from the space of binaryequivalent of number.hence u
get
On Sun, Aug 8, 2010 at 9:37 PM, navin wrote:
> Assunming, integer is 2 byte, What will be the output of the program?
>
> #include
>
> i
sign bit gets extended in case of negative numbers
On Sun, Aug 8, 2010 at 9:37 PM, navin wrote:
> Assunming, integer is 2 byte, What will be the output of the program?
>
> #include
>
> int main()
> {
>printf("%x\n", -1>>1);
>return 0;
> }
>
> [A].
> [B].0fff
> [C].
>
XOR Linked list tutorial http://en.wikipedia.org/wiki/XOR_linked_list
On Aug 8, 11:49 am, Pramod Negi wrote:
> Search XoR List on Wiki
>
> On Sun, Aug 8, 2010 at 10:19 AM, UMESH KUMAR wrote:
>
> > how to convert Doubly Link list to a Singly link list without changes
> > the Structure of the list
@ankur: dp[i][j] number of (prefix) strings that have i As and j Bs is:
dp[i-1][j] number of (prefix) strings that have i-1 As and j Bs appended by
"A"
+
dp[i][j-1] number of (prefix) strings that have i As and j-1 Bs appended by
"B"
@ashish: wikipedia has some nice proofs for catalan number form
Let the problem be represented as f(i,j) for i A's and jB's.
Amir represented it correctly as: f(i,j) = f(i,j-1) + f(j,i-1)
Lets try another representation:
Let t(i,j) is the number of strings of length i+j with iA's and jB's in any
order.
Hence, t(i, j) = (i+j)Ci
Now some valid strings for our p
check out this link guys,,, for more implementation details
On Sun, Aug 8, 2010 at 7:59 PM, ashita dadlani wrote:
> first of all Gene..this is not "he" but "she" :P
> and secondly,can you please mention the details you are talking about so
> that we can reach to a solution?
>
>
> On Sun, Aug 8,
Assunming, integer is 2 byte, What will be the output of the program?
#include
int main()
{
printf("%x\n", -1>>1);
return 0;
}
[A].
[B].0fff
[C].
[D].fff0
Answer: Option A
i dont understand why it produces
actually -1 =
there shoud be one bit zero ( i
if the array is numbered from 0..(2n-1)
i= initial position of int/char
f= final position of int/char
if (i < (2n-1)/2) #for integer
f = i+i
else #for char
f = i - ((2n-1)-i)
so iterate through the array in the following way
choose first element
determine it final position
put the element in th
array:4,5,1,2,3
index: 0,1,2,3,4
real index: 3,4,0,1,2
suppose the index of the least element is k (here k=2)
then when you're applying the binary search the only modification you need
to have is each time you're going to access the ith element of array look at
(i+k)%n th element
--
first find out the element around which array is rotated. i.e find the first
element of original array( using modified binary search )
then compare the element to be searched with the first elent of given array,
if it is greater than it then binary search in first half of array else in
second half
@Manjunath : We can't assume the structure of BST has parent pointers , if
that is explicitly mentioned , then we will have to keep in mind two
pointers one for inorder successor in the left subtree and another for
inorder predecessor in the right subtree and traverse the pointers to find
the
first of all Gene..this is not "he" but "she" :P
and secondly,can you please mention the details you are talking about so
that we can reach to a solution?
On Sun, Aug 8, 2010 at 7:48 PM, Gene wrote:
> Well, then, you must say that in the problem! To to find k, ashita
> dadlani's solution works
Well, then, you must say that in the problem! To to find k, ashita
dadlani's solution works fine, except he has left out important
details. This binary search is a bit tricker that the one to find a
value.
On Aug 8, 2:34 am, Manjunath Manohar wrote:
> hey wait a sec,.. we wont be given the k va
its possible.first apply binary search to find the index where the array is
rotated.
eg.a=[8,9,1,2,3,4,56,7]
first apply binary search to find the value i=2.
now we have 2 sub sorted arrays,ie, from i=0 to i=1 and from i=2 to i=8.
apply binary search to these sub sorted arrays each of which will ta
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