[LeetCode] Single Number II

 Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 

/*
ones is the XOR of all elements that have appeared exactly once so far
twos is the XOR of all elements that have appeared exactly twice so far
Each time we take x to be the next element in the array, there are three cases:

if this is the first time x has appeared, it is XORed into ones
if this is the second time x has appeared, it is taken out of ones (by XORing it again) and XORed into twos
if this is the third time x has appeared, it is taken out of ones and twos.
Therefore, in the end, ones will be the XOR of just one element, the lonely element that is not repeated. There are 5 lines of code that we need to look at to see why this works: the five after x = A[i].

If this is the first time x has appeared, then ones&x=ones so twos remains unchanged. The line ones ^= x; XORs x with ones as claimed. Therefore x is in exactly one of ones and twos, so nothing happens in the last three lines to either ones or twos.

If this is the second time x has appeared, then ones already has x (by the explanation above), so now twos gets it with the line twos |= ones & x;. Also, since ones has x, the line ones ^= x; removes x from ones (because x^x=0). Once again, the last three lines do nothing since exactly one of ones and twos now has x.

If this is the third time x has appeared, then ones does not have x but twos does. So the first line let's twos keep x and the second adds x to ones. Now, since both ones and twos have x, the last three lines remove x from both.
*/
class Solution {
public:
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int ones = 0;
        int twos = 0;
        for(int i = 0; i < n; i++) {
            twos ^= ones&A[i];
            ones ^= A[i];
            int not_three = ~(ones&twos);
            ones &= not_three;
            twos &= not_three;
        }
        return ones;
        
    }
};