672. Bulb Switcher II

There is a room with n bulbs labeled from 1 to n that all are turned on initially, and four buttons on the wall. Each of the four buttons has a different functionality where:

• Button 1: Flips the status of all the bulbs.
• Button 2: Flips the status of all the bulbs with even labels (i.e., 2, 4, …).
• Button 3: Flips the status of all the bulbs with odd labels (i.e., 1, 3, …).
• Button 4: Flips the status of all the bulbs with a label j = 3k + 1 where k = 0, 1, 2, … (i.e., 1, 4, 7, 10, …).

You must make exactly presses button presses in total. For each press, you may pick any of the four buttons to press.

Given the two integers n and presses, return the number of different possible statuses after performing all presses button presses.

class Solution {
    int bitOp(int a, int b) {
        int i = 0, res = 0;
        while(a != 0 or b != 0) {
            int bi = b & 1 ? !(a & 1) : (a & 1);
            res += bi * (1<<i);
            i++;
            a>>=1; b>>=1;
        }
        return res;
    }
public:
    int flipLights(int n, int presses) {
        int bits = (1<<min(n, 3)) - 1;
        int odd = bits & 5, even = bits & 6, over = bits, optional = 1;
        unordered_set<int> res;
        res.insert(bits);
        presses = min(presses, 3);
        while(presses--) {
            unordered_set<int> flip;
            for(auto& bi : res) {
                flip.insert(bitOp(bi,odd));
                flip.insert(bitOp(bi,even));
                flip.insert(bitOp(bi,over));
                if(n >= 3) flip.insert(bitOp(bi,optional));
            }

            swap(res, flip);
        }

        return res.size();
    }
};