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