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
wherek = 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
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):
cnt = mask.bit_count()
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
vis.add(t)
return len(vis)
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);
for (int mask = 0; mask < 1 << 4; ++mask) {
int cnt = Integer.bitCount(mask);
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);
vis.add(t);
}
}
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;
for (int mask = 0; mask < 1 << 4; ++mask) {
int cnt = __builtin_popcount(mask);
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();
}
};
func flipLights(n int, presses int) int {
if n > 6 {
n = 6
}
ops := []int{0b111111, 0b010101, 0b101010, 0b100100}
vis := map[int]bool{}
for mask := 0; mask < 1<<4; mask++ {
cnt := bits.OnesCount(uint(mask))
if cnt <= presses && cnt%2 == presses%2 {
t := 0
for i, op := range ops {
if mask>>i&1 == 1 {
t ^= op
}
}
t &= 1<<6 - 1
t >>= (6 - n)
vis[t] = true
}
}
return len(vis)
}