Given a number, the task is to find XOR of count of 0s and count of 1s in binary representation of a given number.
Input : 5 Output : 3 Binary representation : 101 Count of 0s = 1, Count of 1s = 2 1 XOR 2 = 3. Input : 7 Output : 3 Binary representation : 111 Count of 0s = 0 Count of 1s = 3 0 XOR 3 = 3.
The idea is simple, we traverse through all bits of a number, count 0s and 1s and finally return XOR of two counts.
One observation is, for a number of the form 2^x – 1, the output is always x. We can directly produce answer for this case by first checking n+1 is a power of two or not.
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Improved By : Ita_c