my algo is flawed..i took the matrix to be sorted

On 7/8/10, Ashish Goel <[email protected]> wrote:
> 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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