Maximum Value of XOR in a range?

Question

Given a matrix of size NxN. Rows and Columns are numbered from 0 to N-1. jth column of ith row contains i xor j. In other words, Matrix[i][j] = i ^ j where 0 ? i,j < N. Your task is to find the maximum value occurring in this matrix and the count of its occurrence.

While my approach was

(l..r).each do |i|
    (i..r).each do |j|
      if (i ^ j > max)
        max = i ^ j
      end
    end
  end

I saw this code which says it according to everyone is not quadratic is finding it in linear time.

public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        long N;
        long max    = 1L;
        long count  = 1L;
        long bitN   ;
        StringBuilder sb  = new StringBuilder(4*T);


            line = br.readLine();
            N = Long.parseLong(line);
            {

                bitN    = 1L << getBits(N-1);
                max     = (bitN - 1);
                count   = (bitN == N) ? bitN :(N -(bitN>>1))<<1 ;

                sb.append(max);
                sb.append(" ");
                sb.append(count);
                sb.append("\n");
            }

        System.out.println(sb.toString());
    }

    private static int getBits(long x){
        int count = 0;
        while(x > 0){
            x = x>>1;
            count++;
        }
        return count;
    }
} 

What i'm unable to understand is how is this actually

bitN    = 1L << getBits(N-1);
max     = (bitN - 1);
count   = (bitN == N) ? bitN :(N -(bitN>>1))<<1 ;

able to get the desired result. If anyone of you could just give me basic of this algorithm in simple terms so that i can understand this

Answer 1

Make a table of N versus max and count:

N   max   count   how
1     0       1   0^0
2     1       2   0^1, 1^0
3     3       2   1^2, 2^1
4     3       4   1^2, 2^1, 0^3, 3^0
5     7       2   3^4, 4^3
6     7       4   3^4, 4^3, 2^5, 5^2
7     7       6   3^4, 4^3, 2^5, 5^2, 1^6, 6^1
8     7       8   3^4, 4^3, 2^5, 5^2, 1^6, 6^1, 0^7, 7^0
9    15       2   7^8, 8^7
.
.

The pattern is that, after N crosses a power of two, the max goes up and the count goes back to 2. This is because the bit patterns look like this:

3 = 0b011
4 = 0b100
.
.
7 = 0b0111
8 = 0b1000

bitN is the lowest bit set nowhere in 0..N-1 (so bitN = 8 when N = 8, and bitN = 16 when N = 9). The maximum XOR has all bits below bitN set, which is bitN - 1 by the logic of borrowing in grade-school subtraction. The count goes up two every time N increases by one except when bitN increases as well, when the count resets to two. The purpose of the ternary operator in computing count is to special case N = 1; the left branch is also taken when N is a greater power of two, but the other would work as well.