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Java-based LeetCode algorithm problem solutions, regularly updated
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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`
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