can you clarify ratnesh please
i was thinknig on these lines...
int temp=0; int rowcount=0;
for (int row=0;rownrows;row++)
if (a[row]||temp!=temp)
if (a[row]~0!=0) rowcount++;
for (int col=0;colncols;col++)
if (a[col]||temp!=temp)
if (a[col]~0!=0) colcount++;
Best Regards
Ashish
my algo is flawed..i took the matrix to be sorted
On 7/8/10, Ashish Goel ashg...@gmail.com wrote:
can you clarify ratnesh please
i was thinknig on these lines...
int temp=0; int rowcount=0;
for (int row=0;rownrows;row++)
if (a[row]||temp!=temp)
if (a[row]~0!=0) rowcount++;
for (int
for(i=0 to n-1)
if( binarysearch(i,n-1,1) + 1)
count++
print count.
binarysearch(first,last,item)
if(1 is there)
return mid
else
return -1.
similarly we can go for coloumns.
o(nlogn)
On 7/5/10, divya jain sweetdivya@gmail.com wrote:
i think u need to visit every element atleast
For a given matrix NxN having 0 or 1’s only. You have to find the
count of rows and columns where atleast one 1 occurs.
e,g
0 0 0 0
1 0 0 1
1 0 0 1
1 1 0 1
Row count having 1 atleast once: 3
Col count having 1 atleast once: 3
Any Solution less than O(n^2) will do
--
You received
i think u need to visit every element atleast once to see if its 1 or 0, nd
so update the count.
so i dont think it will be possible in less than O(n2)
On 5 July 2010 15:41, amit amitjaspal...@gmail.com wrote:
For a given matrix NxN having 0 or 1’s only. You have to find the
count of rows
