Given two integers A and B, the task is to find an integer X such that (X XOR A) is minimum possible and the count of set bit in X is equal to the count of set bits in B.
Input: A = 3, B = 5
Binary(A) = Binary(3) = 011
Binary(B) = Binary(5) = 101
The XOR will be minimum when M = 3
i.e. (3 XOR 3) = 0 and the number
of set bits in 3 is equal
to the number of set bits in 5.
Input: A = 7, B = 12
Approach: It is known that the xor of an element with itself is 0. So, try to generate M’s binary representation as close to A as possible. Traverse from the most significant bit in A to the least significant bit and if a bit is set at the current position then it also needs to be set in the required number in order to minimize the XOR but the number of bits set has to be equal to the number of set bits in B. So, when the count of set bits in the required number has reached the count of set bits in B then the rest of the bits have to be 0.
Below is the implementation of the above approach:
Time Complexity: O(log(N))
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