anybody have informaton regarding questions asked in written and
interview of capillary technology for developer post
please share at bhardwaj.ankit...@gmail.com
thanks in advance.
On 5/22/12, Navin.nitjsr wrote:
> If the matrix is 4-connected, we can use the same matrix.
> now we have to find
@praveen : little more clarity required in your algoare you calling it
recursively or moving row by row.
On Tue, Jan 24, 2012 at 5:59 PM, praveen raj wrote:
> Idea:
> 1)Take count =0;
> 2) make Outer loop ...and search for 1's .
> 3) Start ...searching for 1 consecutively...
> and ma
name it.
PRAVEEN RAJ
DELHI COLLEGE OF ENGINEERING
On Wed, Jan 25, 2012 at 12:45 AM, atul anand wrote:
> @Praveen : i have doubt in your algo...it seem it may fail for some
> cases...
>
> On Tue, Jan 24, 2012 at 5:59 PM, praveen raj wrote:
>
>> Idea:
>> 1)Take count =0;
>> 2) make Outer loo
@Praveen : i have doubt in your algo...it seem it may fail for some cases...
On Tue, Jan 24, 2012 at 5:59 PM, praveen raj wrote:
> Idea:
> 1)Take count =0;
> 2) make Outer loop ...and search for 1's .
> 3) Start ...searching for 1 consecutively...
> and make it ..0 untill all consecutive
Idea:
1)Take count =0;
2) make Outer loop ...and search for 1's .
3) Start ...searching for 1 consecutively...
and make it ..0 untill all consecutive 1's becomes 0..
and then count++
4) go to 1) untill all 1's finished..
count will give the total number of islands...
PRAVEEN RAJ
DE
@Umer : will fail for this case :-
0 0 0 0
0 1 0 1
1 1 1 1
island = 1;
it seem your code will print island =1 ;
On Sun, Jan 15, 2012 at 3:45 PM, Umer Farooq wrote:
> Here is my solution. Please have a look at it. Any kind of positive
> criticism will be highly appreciated.
>
> bool isConnecte
Here is my solution. Please have a look at it. Any kind of positive
criticism will be highly appreciated.
bool isConnected(int **space, int x, int y)
{
if (x == 0 && y == 0)
{
return false;
}
if (y > 0)
{
if (space[x][y-1] == 1)
return true;
}
if (x > 0)
{
if (space[x-1][y] == 1)
return true;
}
if
this is the solution that i was referring to in the link i provided.
On the same lines there is another problem of rat in a maze .
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Wed, Jan 11, 2012 at 7:19 AM, Gene wrote:
> Guys,
>
> You are ma
@Umer : it has 1 island ashish made editing mistake before.
On Wed, Jan 11, 2012 at 11:58 AM, Umer Farooq wrote:
> I still don't get how are they two islands. As long as I have understood,
> diagonals abridge the two islands into one. In this case, these two islands
> are connected so they
I still don't get how are they two islands. As long as I have understood,
diagonals abridge the two islands into one. In this case, these two islands
are connected so they form one single island?
1 1 0 0
1 1 0 0
0 0 1 1
This can be either one single island. Or they are three island if a change
in
I think atul/Ramakanth's approach will work fine, if we include one more
condition
for each arr[i][j]
if(arr[i][j]==1)
{
if (arr[i-1][j]==0 && arr[i][j-1]==0 && arr[i-1][j-1]==0)
count++;
else if (arr[i-1][j]==1 && arr[i][j-1]==1 && arr[i-1][j-1]==0)
count--;
}
On Wed, Jan 11, 2012 at 8:10 AM, s
@gene
in that case ur erase() should even consider diagonal elements as well,
else there would be 2 islands in example
surender
On Wed, Jan 11, 2012 at 7:19 AM, Gene wrote:
> Guys,
>
> You are making this way too hard. It's really a graph problem. The
> nodes are the 1's and adjacent 1's are c
while all this is fine, the basic test case that each point on the circle is
at a distance of r from the centre becomes first functional test case.
what would be non-functional test cases eg. to check on different
dpi/screen sizes etc
Best Regards
Ashish Goel
"Think positive and find fuel in fai
13 matches
Mail list logo
