How is XOR able to give unique number from multiple inputs?

#1

from given test case, i.e. 3, 2, 3, 5, 6,5, 2

ans = 0;

1st run : 0^3 = 3;
2nd run: 3^2 = 1;
3rd run: 1^3 = 2;
4th run: 2^5 = 7;
5th run: 7^6 = 1;
6th run: 1^5 = 4;
7th run: 4^2 = 6;

In the end, we came to the resultant answer, I get that. But I don’t understand why is if intermediate ans are different and it doesn’t even know if the previous number was repeated.

Suppose there are 6 digits only, the output is outside the input given, the output is XOR of remaining two unique numbers.

I am unable to understand the logic beyond two same numbers give 0 as output for their XOR.

#2

This is because XOR only cares about bits. So intermediate state can be anything, but when you add a number x, bits corresponding to it get added. So when you again add x, buts corresponding to it will get removed, no matter what.

#3

so XOR can only be used for finding unique number in the array and not to check how many numbers are occuring only once?

#4

Yes it only finds the unique number if all the other numbers have even occurences.

#5

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