can you clarify ratnesh please
i was thinknig on these lines...
int temp=0; int rowcount=0;
for (int row=0;row<nrows;row++)
if (a[row]||temp!=temp)
if (a[row]&~0!=0) rowcount++;
for (int col=0;col<ncols;col++)
if (a[col]||temp!=temp)
if (a[col]&~0!=0) colcount++;
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Wed, Jul 7, 2010 at 9:53 PM, Ratnesh Thakur <[email protected]>wrote:
> 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 <[email protected]> wrote:
> > 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 <[email protected]> wrote:
> >
> >>
> >>
> >> 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....
> >>
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