As the array elements are previously sorted ,we will check for middle elements of the array only, i.e:
If n is odd then we will check for (n/2)th element (element at the center) and if it is 0 then 0 is in majority , otherwise 1 is in majority.
if n is even then we will first check for (n/2)th element and if it is 0 then 0 is in majority and if it is 1 then we will check for (n/2-1)th element , if it is 1 then 1 is in majority.
In both cases number of comparison required are of O(1)
Hence 1) is the correct answer...