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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0672_bulb_switcher_ii;

// #Medium #Math #Depth_First_Search #Breadth_First_Search #Bit_Manipulation
// #2022_03_22_Time_0_ms_(100.00%)_Space_38.9_MB_(80.85%)

/**
 * 672 - Bulb Switcher II\.
 *
 * Medium
 *
 * 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`
**/
public 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;
        }
    }
}