672. 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:

Flip all the lights.
Flip lights with even numbers.
Flip lights with odd numbers.
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].

 

We only need to consider special cases which n<=2 and m < 3. When n >2 and m >=3, the result is 8.
The four buttons:

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, ...

If we use button 1 and 2, it equals to use button 3.
Similarly...

1 + 2 --> 3, 1 + 3 --> 2, 2 + 3 --> 1
So, there are only 8 cases.

All_on, 1, 2, 3, 4, 1+4, 2+4, 3+4

And we can get all the cases, when n>2 and m>=3.

class Solution {
    public int flipLights(int n, int m) {
        if (m == 0) return 1;
        if (n <= 0 || m < 0) return 0;
        
        if (n == 1) return 2;
        else if (n == 2) return (m == 1) ? 3 : 4;
        else return (m == 1) ? 4 : ((m == 2) ? 7 : 8);
    }
}

　　

1: light is on
0: light is off

n == 1

Only 2 possibilities: 1 and 0.

n == 2

After one operation, it has only 3 possibilities: 00, 10 and 01.
After two and more operations, it has only 4 possibilities: 11, 10, 01 and 00.

n == 3

After one operation, it has only 4 possibilities: 000, 101, 010 and 011.
After two operations, it has 7 possibilities: 111,101,010,100,000,001 and 110.
After three and more operations, it has 8 possibilities, plus 011 on above case.

n >= 4

After one operation, it has only 4 possibilities: 0000, 1010, 0101 and 0110.
After two or more operations: it has 8 possibilities, 1111,1010,0101,0111,0000,0011, 1100 and 1001.