@Ramakant : i guess you shud include diagonal case also ....

for each arr[i][j]
if(arr[i][j]==1)
{
if (!(arr[i-1][j]==1 || arr[i][j-1]==1 || arr[i-1][j-1]))
count++;
}




On Tue, Jan 10, 2012 at 9:33 AM, Ramakant Sharma <ramakant...@gmail.com>wrote:

> Scan the matrix row wise left to right........
> for each arr[i][j]
> if(arr[i][j]==1)
> {
> if (!(arr[i-1][j]==1||arr[i][j-1]==1))
> count++;
> }
> ///also chk for baundary values accordingly
>
>
>
>
> 1 1 0 0
> 1 1 0 0
> 0 0 1 1
>
>
> i think it should work..
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> To unsubscribe from this group, send email to
> algogeeks+unsubscr...@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>

-- 
You received this message because you are subscribed to the Google Groups 
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to 
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at 
http://groups.google.com/group/algogeeks?hl=en.

Reply via email to