@Dave
Slight change u have to do
#includestdio.h
main()
{
int a = 24;
int b = (a | 7)+1; //This Line U have to change. not a || 7.
printf(%d,b);
return 0;
}
tats it
btw it is nice one line easy soln...
On Sat, Sep 25, 2010 at 10:17 PM, Dave dave_and_da...@juno.com wrote:
@coolfrog$
I think this can match ur requirements not even indirect recursion
#includeiostream
using namespace std;
static int i;
class sample
{
public:
sample()
{
i=i+1;
couti\n;
}
};
main()
{
int n;
coutEnter n\n;
cinn;
sample s[n];
return 0;
}
tats it
On Thu, Sep 23, 2010 at 11:29 PM,
BSTP : Root's right and left subtrees are BST and value at Root is
(greater than largest of left) and (smaller than lowest of right).
if BSTP is true, size of this BST is sum of (size of left subtree) and
(size of right subtree) plus 1. Compare this value with global
maximum.
Do it recursively.
This can also be done if we do an inorder traversal of the binary tree and
look for the longest continuous sequence of numbers in ascending order.
On Sun, Sep 26, 2010 at 11:10 AM, mac adobe macatad...@gmail.com wrote:
No parody .. that would be another doubt :(
On Sat, Sep 25, 2010 at
@Mohit:
I dont think it really matters here.We just have to validate the snapshot of
the game board.Number of players should not have any relevance here.
On Sat, Sep 25, 2010 at 2:46 PM, mohit ranjan shoonya.mo...@gmail.comwrote:
@Ashita,
Your logic is fine for one vs one game, but as per
@ ashita and vrinda
can u please write ur ms written and interview questions..
i ll be really thankful to both of u as ms is going to visit my campus
soon.. so plz help...
On Sep 26, 1:58 pm, ashita dadlani ash@gmail.com wrote:
@Mohit:
I dont think it really matters here.We just have to
sry...i dnt get the qstn clearly.
On Sat, Sep 25, 2010 at 10:18 PM, Dave dave_and_da...@juno.com wrote:
Simpler: answer = (x || 7) + 1
Dave
On Sep 25, 10:14 am, rajess rajess1...@yahoo.com wrote:
scan last(left) 4 bits .if the bits are not 1000.make them 1000.else
scan from left
@Dave
Your answer will be always 2 irrespective of the value of 'x'.
BTW I too didn't the question.
On Sep 26, 2:42 pm, Mahendran MaheM mehamind...@gmail.com wrote:
sry...i dnt get the qstn clearly.
On Sat, Sep 25, 2010 at 10:18 PM, Dave dave_and_da...@juno.com wrote:
Simpler: answer
@Shrevan: I mistyped what I intended. Try answer = (x | 7) + 1;
Dave
On Sep 26, 5:51 am, Shravan shravann1...@gmail.com wrote:
@Dave
Your answer will be always 2 irrespective of the value of 'x'.
BTW I too didn't the question.
On Sep 26, 2:42 pm, Mahendran MaheM mehamind...@gmail.com wrote:
@Ashwath: Thanks for the correction.
Dave
On Sep 26, 1:20 am, aswath G B aswat...@gmail.com wrote:
@Dave
Slight change u have to do
#includestdio.h
main()
{
int a = 24;
int b = (a | 7)+1; //This Line U have to change. not a || 7.
printf(%d,b);
return 0;
}
tats
On Tue, Sep 21, 2010 at 6:05 PM, saurabh singh saurabh.n...@gmail.comwrote:
solution 1:
use concept of quad-tree and do binary search in that tree
solution 2:
do binary search on major diagonal. ultimately u will narrow down to search
for element in 2 rows. in these two rows again do
Here's an O(n) soln:
Start from the bottom right corner. Move up within the last column until you
reach an element such that the element before that is less than the value
being searched and this is greater.
Now move left and check for the same.
Move one more left. The value can only be in the
Sorry the last solution wont work.
Here's the correct O(n) soln:
Start from top-right element.
If it is greater than the item, go left.
Else go down.
Keep on doing this until you find the element or can not go left or down
(then the element is not in the array).
void 2dsearch(int i, int j,
As you mentioned ultimately element to be searched should either be in row
'i' (ahead of [i,i] element) or in row i+1 (before [i+1,i+1] element). Since
each row contain numbers in sorted order so u can do binary search on these
two rows and ultimately the complexity will be O(logn) only
On Sun,
15
/\
8 25
/\
20 22
On Sep 26, 10:45 am, Chonku cho...@gmail.com wrote:
This can also be done if we do an inorder traversal of the binary tree and
look for the longest continuous
If the array is m by n, pick the element a[m/2][n/2], i.e. the one in
the middle. There are now 3 possibilities:
1) The middle element is the one you're looking for, so you're done.
2) The element you're looking for is smaller. In this case you can
throw out about 1/4 of the array: everything
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