672. Bulb Switcher II

Problem

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

Solution (Java)

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).