Given an integer A, the task is to find the maximum value possible(B) which is less than A, such that xor of these two numbers A and B is equal their sum, that is A ^ B = A + B.
Input: A = 4
There are many such integers such that A ^ B = A + B
Some of this integers are –
4 ^ 3 = 4 + 3 = 7
4 ^ 2 = 4 + 2 = 6
4 ^ 1 = 4 + 1 = 5
4 ^ 0 = 4 + 0 = 4
Maximum of these values is 3
There is no integer except 0 such that A + B = A ^ B
Approach: The idea is to use the fact that and to get the value of , the value of (A & B) must be equal to 0.
=> A & B = 0 => B = ~A
A = 4 (1 0 0) B = ~ A = (0 1 1) = 3
Below is the implementation of the above approach:
- Time Complexity: In the above-given approach, there is conversion from decimal to binary which takes O(logN) time in worst case. Therefore, the time complexity for this approach will be O(logN).
- Auxillary Space Complexity: In the above-given approach, there is no extra space used. Therefore, the auxillary space complexity for the above approach will be O(1)
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