superb answer by dave
On Sun, May 15, 2011 at 8:27 AM, Amol Sharma wrote:
> like utkarsh answer
> --
>
>
> Amol Sharma
> Second Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>
>
>
>
> On Wed, May 11, 2011 at 12:58 AM, Dave wrote:
>
>> 19 = XIX in Roman Numerals. Take 1 = I
like utkarsh answer
--
Amol Sharma
Second Year Student
Computer Science and Engineering
MNNIT Allahabad
On Wed, May 11, 2011 at 12:58 AM, Dave wrote:
> 19 = XIX in Roman Numerals. Take 1 = I away and you have XX = 20.
>
> Dave
>
> On May 10, 2:09 am, Lavesh Rawat wrote:
> > *MATHS TRICK TE
@Pacific: A balanced binary tree is commonly defined as a binary tree
in which the height of the two subtrees of every node never differ by
more than 1. Thus, there could be more nodes in one subtree than in
the other. E.g., a balanced binary tree with 11 nodes could have 7
nodes in the left subtre
> After expanding B, the border is with G (25)
>
> Node G is the smallest of the border.
>
> You finish the algorithm when the node G is the smallest of the border not
> when he enters the boundary
Yes, we finish with G=25, but we still have G=35 first.
This is all right, since I understand now th
i think we can use heaps for this problem..bring to the root which has
capacity to hold and pick the root each time. If the root cannot accomodate
then no other node will be able to accomodate.
On Sat, May 14, 2011 at 1:26 AM, MK wrote:
> Consider the first fit strategy for online bin packing.
>
Will not a balanced binary tree do the job ? if we will pick the root each
time for the median.
On Sat, May 14, 2011 at 9:10 PM, Dave wrote:
> @Ashish: The idea is to keep two heaps, a max-heap of the smallest
> half of the elements and a min-heap of the largest elements. You
> insert incoming
On May 13, 1:24 am, amit wrote:
> Given an array A[i..j] find out maximum j-i such that A[i] O(n) time.#include
using namespace std;
#define max 12
int find(int a[],int i,int j)
{
while(i=j)
return 0;
}
int main()
{
int a[max]={ 18,20,43,2,33,45,32,55,4,33,22,34
@Ashish: The idea is to keep two heaps, a max-heap of the smallest
half of the elements and a min-heap of the largest elements. You
insert incoming elements into the appropriate heap. If the result is
that the number of elements in the two heaps differs by more than 1,
then you move the top element
Each team plays a total of 14 matches. Top 50% teams(4 out of 8) qualify for
the semis.
Thus, u must win more than 50% matches to be sure to get into semis. Thus, 8
is the answer.
On Fri, May 13, 2011 at 12:14 AM, amit wrote:
> Consider a series in which 8 teams are participating. each team play
not clear, can u elaborate..
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Fri, May 13, 2011 at 7:15 PM, Bhavesh agrawal wrote:
> This problem can be solved using 2 heaps and the median can always be
> accessed in O(1) time ,the first node.
>
@all i have posted the solution of same problem few times back ,
search in group thread
i used BST & using that inversion count can be calculated in O(nlogn)
if you found any error on that then let me know
Thanks
Shashank
CSE,BIT Mesra
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