Given two positive integers A and B such that A != B, the task is to find a positive integer X which maximizes the expression (A AND X) * (B AND X).
Input: A = 9 B = 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)
Input: A = 11 and B = 13
Naive approach: One can run a loop from 1 to max(A, B) and can easily find X which maximizes the given expression.
Efficient approach: It is known that,
(a – b)2 ≥ 0
which implies (a + b)2 – 4*a*b ≥ 0
which implies a * b ≤ (a + b)2 / 4
Hence, it concludes that a * b will be maximum when a * b = (a + b)2 / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)
Now X can be found as:
A = 11 = 1011
B = 13 = 1101
X = ? = abcd
At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1
Hence, X = 1001 = 9
Below is the implementation of the above approach:
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