# 672. Bulb Switcher II (Medium)

https://leetcode.com/problems/bulb-switcher-ii/

There is a room with `n` lights which are turned on initially and 4 buttons on the wall. After performing exactly `m` unknown operations towards buttons, you need to return how many different kinds of status of the `n` lights could be.

Suppose `n` lights are labeled as number [1, 2, 3 ..., n], function of these 4 buttons are given below:

1. Flip all the lights.
2. Flip lights with even numbers.
3. Flip lights with odd numbers.
4. Flip lights with (3k + 1) numbers, k = 0, 1, 2, ...

Example 1:

```Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]
```

Example 2:

```Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]
```

Example 3:

```Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].
```

Note: `n` and `m` both fit in range [0, 1000].

## Solutions

``````class Solution {
// For same operations that performed odd times, the outcome is same with 1 time.
// For same operations that performed even times, the outcome is same with 0 time.

// The effect of flipping button 1 equates with flipping both button 2 and 3.
// Literally, among button 1, 2, 3, each one can be expressed with other two.
// Consequently, we have at most 8 possible combinations for these four kinds of buttons

// Specially,  if n=1, button 2 is invalid, and button 1, 3, 4 are equivalent;
// if n=2, button 4 equates with button 3.
public int flipLights(int n, int m) {
if (n == 0 || m == 0) {
return 1;
}

if (n == 1 && m >= 1) {
return 2;
}

if (n == 2) {
if (m == 1) {
return 3;
}

if (m >= 2) {
return 4;
}
}

if (n >= 3) {
if (m == 1) {
return 4;
}

if (m == 2) {
return 7;
}
}

return 8;
}
}
``````