Given two positive integers A and B. Let’s define D such that B AND D = D. The task is to maximize the expression A XOR D.
Input: A = 11 B = 4 Output: 15 Take D = 4 as (B AND D) = (4 AND 4) = 4. Also, (A XOR D) = (11 XOR 4) = 15 which is the maximum according to the given condition. Input: A = 9 and B = 13 Output: 13
Naive approach: Since B AND D = D, D will always be smaller than or equal to B. Hence, one can run a loop from 1 to B and check whether the given conditions are satisfied or not.
Efficient approach: Instead of running a loop and checking for each D, the maximum value of the expression (A XOR D) can be easily calculated using Bit Manipulation techniques.
Let’s take an example to understand the way to approach the problem:
A = 11 = 1011, B = 14 = 1110 Let's assume D = abcd in base 2 notation B AND D: 1110 A XOR D: 1011 abcd abcd ------ ------ abcd ???? At 0th place: (0 AND d) = d implies d = 0 At 1st place: (1 AND c) = c implies c = 0, 1 but to maximize (A XOR D), take c = 0 At 2nd place: (1 AND b) = b implies b = 0, 1 but to maximize (A XOR D), take b = 1 At 3rd place: (1 AND a) = a implies a = 0, 1 but to maximize (A XOR D), take a = 0 Hence, D = 0100 = 4 and maximum value of (A XOR D) = (11 XOR 4) = 15.
Below is the implementation of the above approach:
- Array Manipulation and Sum
- Bit manipulation | Swap Endianness of a number
- Bits manipulation (Important tactics)
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