Instead of having to test various combinations of A & B, once can simply determine the result with knowledge of the value of the last digit of k.

Thought Process: Always strive to see if k-1 can be achieved, if not then try k-2 and so on. This is because the maximum value less than k is k-1, but (k-1) will only be valid if it can be formed via A&B

Here it goes:

(I) If k ends with 1, then the result is (k-1) In this case k-1 can always be formed by switching off the last digit of k, where k is .......1

(II) If k ends with 0, and is of form 2^m i.e. a power of 2 then result is (k-1) or (k-2). Let k=2^m If (2^{m+1})-1 is in the range, then result is (k-1). In this case B=(2^{m+1})-1 and A=k-1

else, result is (k-2). (k-2) is always guaranteed as it is the precursor of a binary number that ends with a 1, see (I)

(III) The last possible case is where k ends with 0, and is not a power of 2. Again the result is either (k-1) or (k-2)

If the number obtained by switching on the rightmost 0 of (k-1) is the range, then that number can be bitwise multiplied by k-1 to give (k-1). Otherwise result is (k-2). (k-2) is always guaranteed as it is the precursor of a binary number that ends with a 1, see (I) & (II)

see code: https://github.com/niranfor1/superQuantHackerrank/blob/master/30DaysOfCodeBitwiseAND.cpp