672. Bulb Switcher II

Description

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.

Example 1:

Input: n = 1, presses = 1
Output: 2
Explanation: Status can be:
- [off] by pressing button 1
- [on] by pressing button 2


Example 2:

Input: n = 2, presses = 1
Output: 3
Explanation: Status can be:
- [off, off] by pressing button 1
- [on, off] by pressing button 2
- [off, on] by pressing button 3


Example 3:

Input: n = 3, presses = 1
Output: 4
Explanation: Status can be:
- [off, off, off] by pressing button 1
- [off, on, off] by pressing button 2
- [on, off, on] by pressing button 3
- [off, on, on] by pressing button 4


Constraints:

• 1 <= n <= 1000
• 0 <= presses <= 1000

Solutions

• class Solution {
public int flipLights(int n, int presses) {
int[] ops = new int[] {0b111111, 0b010101, 0b101010, 0b100100};
Set<Integer> vis = new HashSet<>();
n = Math.min(n, 6);
if (cnt <= presses && cnt % 2 == presses % 2) {
int t = 0;
for (int i = 0; i < 4; ++i) {
if (((mask >> i) & 1) == 1) {
t ^= ops[i];
}
}
t &= ((1 << 6) - 1);
t >>= (6 - n);
}
}
return vis.size();
}
}

• class Solution {
public:
int flipLights(int n, int presses) {
n = min(n, 6);
vector<int> ops = {0b111111, 0b010101, 0b101010, 0b100100};
unordered_set<int> vis;
if (cnt > presses || cnt % 2 != presses % 2) continue;
int t = 0;
for (int i = 0; i < 4; ++i) {
if (mask >> i & 1) {
t ^= ops[i];
}
}
t &= (1 << 6) - 1;
t >>= (6 - n);
vis.insert(t);
}
return vis.size();
}
};

• class Solution:
def flipLights(self, n: int, presses: int) -> int:
ops = (0b111111, 0b010101, 0b101010, 0b100100)
n = min(n, 6)
vis = set()
for mask in range(1 << 4):
if cnt <= presses and cnt % 2 == presses % 2:
t = 0
for i, op in enumerate(ops):
if (mask >> i) & 1:
t ^= op
t &= (1 << 6) - 1
t >>= 6 - n
return len(vis)


• func flipLights(n int, presses int) int {
if n > 6 {
n = 6
}
ops := []int{0b111111, 0b010101, 0b101010, 0b100100}
vis := map[int]bool{}
if cnt <= presses && cnt%2 == presses%2 {
t := 0
for i, op := range ops {
t ^= op
}
}
t &= 1<<6 - 1
t >>= (6 - n)
vis[t] = true
}
}
return len(vis)
}