Single Number III — C++

// Time:  O(n)
// Space: O(1)

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        // Xor all the elements to get x ^ y.
        const auto x_xor_y = accumulate(nums.cbegin(), nums.cend(), 0u, bit_xor<int>());

        // Get the last bit where 1 occurs by "x & ~(x - 1)"
        // Because -(x - 1) = ~(x - 1) + 1 <=> -x = ~(x - 1) 
        // So we can also get the last bit where 1 occurs by "x & -x"
        const auto bit = x_xor_y & -x_xor_y;

        // Get the subset of A where the number has the bit.
        // The subset only contains one of the two integers, call it x.
        // Xor all the elements in the subset to get x.
        vector<int> result(2, 0);
        for (const auto& i : nums) {
            result[static_cast<bool>(i & bit)] ^= i;
        }
        return result;
    }
};

// Time:  O(n)
// Space: O(1)
class Solution2 {
public:
    vector<int> singleNumber(vector<int>& nums) {
        // Xor all the elements to get x ^ y.
        uint32_t x_xor_y = 0;
        for (const auto& i : nums) {
            x_xor_y ^= i;
        }

        // Get the last bit where 1 occurs.
        const auto bit = x_xor_y & ~(x_xor_y - 1);

        // Get the subset of A where the number has the bit.
        // The subset only contains one of the two integers, call it x.
        // Xor all the elements in the subset to get x.
        uint32_t x = 0;
        for (const auto& i : nums) {
            if (i & bit) {
                x ^= i;
            }
        }

        return {int(x), int(x_xor_y ^ x)};
    }
};