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 navin.nit...@gmail.com wrote:
If the matrix is 4-connected, we can use the same matrix.
If the matrix is 4-connected, we can use the same matrix.
now we have to find the total number of connected components of a graph.
consider
1 1 1 0 0 1 1 0 1
0 1 1 1 0 0 1 0 0
1 1 0 1 0 1 1 1 0
0 0 0 0 0 0 0 0 1
we can use bfs/dfs to mark the nodes as visited and thus total
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
@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 praveen0...@gmail.com 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
name it.
PRAVEEN RAJ
DELHI COLLEGE OF ENGINEERING
On Wed, Jan 25, 2012 at 12:45 AM, atul anand atul.87fri...@gmail.comwrote:
@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 praveen0...@gmail.comwrote:
Idea:
@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 praveen0...@gmail.com wrote:
Idea:
1)Take count =0;
2) make Outer loop ...and search for 1's .
3) Start ...searching for 1
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 (x
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 gene.ress...@gmail.com wrote:
Part of the problem must be rules that specify how 1's can be
connected to form an island.
From the discussion, it looks like the rules are that a 1 must be
connected North, West, East, or South. This is called a 4-connected
grid.
Another possibility would be an 8-connected grid. This is
The OP is not clear on how to handle diagonals. If adjacent 1's on the
diagonal are considered connected, then just add 4 more calls to
erase().
The standard terms are 4-connected and 8-connected. Both come up when
working with grid graphs or pixel matrices.
On Jan 10, 9:40 pm, surender sanke
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 connected by undirected edges.
You must count components in the graph. So the algorithm is easy:
Find a component, erase it, repeat. Count components as you go.
What's an efficient
@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 gene.ress...@gmail.com wrote:
Guys,
You are making this way too hard. It's really a graph problem. The
nodes are the 1's and
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, surender
@Umer : it has 1 island ashish made editing mistake before.
On Wed, Jan 11, 2012 at 11:58 AM, Umer Farooq the.um...@gmail.com 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
@harshal every thing seems to be correct m not sure if things will
mess up but can we do it in 1 pass ..??
Shashank
CSE,BIT Mesra
Cracking The Code shashank7s.blogspot.com
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@ross i think u mean s[i]==g[b[i]] or s[i]==g[i] then hit++ isn't it
u r not using guess at all as u r comparing character with digit
correct me if m wrong
Shashank
CSE,BIT Mesra
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An algorithm is:
Have a bit array B of 256/65k size.
If an color 'i' is encountered in the solution, set its B[i]=1;
Traverse the solution array S,
if(S[i]==B[i]) hits++;
else if ( B[S[i]] ) pseudohits++;
On Jun 10, 9:40 am, Harshal
Are we drawing a circle on the screen?
In addition to Harshal's suggestions, try putting the center off the
screen, but have part of the circle on the screen:
x=-10
y=-20
r=100
Don
On Jun 2, 9:19 am, Ashish Goel ashg...@gmail.com wrote:
Given a function to draw a circle with input paramters
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
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