# Question

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].

# Solution

As before, the first 6 lights uniquely determine the rest of the lights. This is because every operation that modifies the x-th light also modifies the $$(x+6)$$-th light, so the x-th light is always equal to the $$(x+6)$$-th light.

Actually, the first 3 lights uniquely determine the rest of the sequence, as shown by the table below for performing the operations a, b, c, d:

• Light 1 = 1 + a + c + d
• Light 2 = 1 + a + b
• Light 3 = 1 + a + c
• Light 4 = 1 + a + b + d
• Light 5 = 1 + a + c
• Light 6 = 1 + a + b

So that (modulo 2):

• Light 4 = (Light 1) + (Light 2) + (Light 3)
• Light 5 = Light 3
• Light 6 = Light 2

The above justifies taking $$n = min(n, 3)$$ without loss of generality. The rest is now casework.

Let’s denote the state of lights by the tuple $$(a, b, c)$$. The transitions are to XOR by (1, 1, 1), (0, 1, 0), (1, 0, 1), or (1, 0, 0).

When m = 0, all the lights are on, and there is only one state (1, 1, 1). The answer in this case is always 1.

When m = 1, we could get states (0, 0, 0), (1, 0, 1), (0, 1, 0)(, or (0, 1, 1). The answer in this case is either 2, 3, 4 for n = 1, 2, 3 respectively.

When m = 2, we can manually check that we can get 7 states: all of them except for (0, 1, 1). The answer in this case is either 2, 4, 7 for n = 1, 2, 3 respectively.

When m = 3, we can get all 8 states. The answer in this case is either 2, 4, 8 for n = 1, 2, 3 respectively.

class Solution(object):
def flipLights(self, n, m):
"""
:type n: int
:type m: int
:rtype: int
"""
n = min(n, 3)
if m == 0: return 1
if m == 1: return [2, 3, 4][n-1]
if m == 2: return [2, 4, 7][n-1]
return [2, 4, 8][n-1]