Bulb Switcher II

Medium

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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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints for `n` (number of bulbs) and `presses`?
If `presses` is 0, should I return 1 (all bulbs on) or is there another expected behavior?
Are `n` and `presses` guaranteed to be non-negative integers?
The problem mentions 'different possible states'. Does the order in which the buttons are pressed matter, or only the total number of times each button is effectively pressed?
Are we only concerned with the final state of the bulbs, or do we need to track the intermediate states during the button presses?

Brute Force Solution

Approach

The brute force approach for the bulb switcher problem involves trying every possible combination of button presses. We'll explore all the different sequences of button presses and see how they affect the lights. The goal is to count how many distinct final states of the lights are possible.

Here's how the algorithm would work step-by-step:

List all possible combinations of pressing the four buttons.
For each combination, figure out which lights are on and which are off after pressing those buttons.
Keep track of the different arrangements of lights (on and off) that you find.
If you see the same arrangement of lights more than once from different button combinations, only count it once.
Count the total number of different light arrangements you found. This is the answer.

Code Implementation

def flip_lights_brute_force(number_of_bulbs, number_of_presses):
    possible_states = set()

    for i in range(1 << (number_of_presses * 1)):
        # Generate all possible combinations of button presses
        button_presses = []
        for j in range(number_of_presses):
            if (i >> j) & 1:
                button_presses.append(1)
            else:
                button_presses.append(0)

        bulbs = [1] * number_of_bulbs

        # Simulate the effect of pressing the buttons
        for k in range(number_of_presses):
            if button_presses[k]:
                if k == 0:
                    for index in range(number_of_bulbs):
                        bulbs[index] = 1 - bulbs[index]

                elif k == 1:
                    for index in range(0, number_of_bulbs, 2):
                        bulbs[index] = 1 - bulbs[index]

                elif k == 2:
                    for index in range(1, number_of_bulbs, 2):
                        bulbs[index] = 1 - bulbs[index]

                else:
                    for index in range(0, number_of_bulbs, 3):
                        bulbs[index] = 1 - bulbs[index]

        # Record the final state of the bulbs as a tuple
        possible_states.add(tuple(bulbs))

    # Remove any duplicate bulb states, then return the result
    return len(possible_states)

Big(O) Analysis

Time Complexity

O(2^m * n) – The algorithm iterates through all possible combinations of pressing m buttons. There are 2^m such combinations since each button can be either pressed or not pressed. For each button combination, the algorithm iterates through all n bulbs to determine the final state. Therefore, the time complexity is O(2^m * n), where m is the number of buttons and n is the number of bulbs. Since the number of buttons is fixed at 4, this can be simplified to O(n).

Space Complexity

O(2^min(n,3)) – The brute force approach stores distinct arrangements of lights. The number of distinct arrangements depends on the number of bulbs, n. Each bulb can be either on or off, leading to potentially 2^n distinct arrangements. The plain English explanation explicitly mentions keeping track of arrangements of lights and only counting unique ones. The number of possible configurations of the lights can be further reduced by considering the overlapping effect of buttons and that the result is capped to 8 configurations when n >= 3. Therefore, the space used is proportional to 2^min(n,3), where n is the number of bulbs.

Optimal Solution

Approach

The trick to solving this problem efficiently is to realize that the number of bulbs and button presses don't matter as much as they seem. There are only a limited number of different states you can reach, so we can figure out what all the possible final configurations of the bulbs are without testing every single combination.

Here's how the algorithm would work step-by-step:

Notice that after 3 bulbs, the pattern of possible on/off states repeats.
Figure out all the different combinations you can get with the button presses for a small number of bulbs (like 1, 2, and 3).
Consider the constraint on the number of button presses available. Fewer presses limits the states we can reach.
Create a set of possible states. Each element represents a unique configuration of on/off bulbs that can be achieved.
If the number of bulbs is small (1, 2 or 3) find all the possible states explicitly from first principles.
Return the number of possible different states.
The number of possible states depends on the number of bulbs and number of button presses.
Handle cases where the maximum number of button presses is zero, which leads to exactly one possible state.

Code Implementation

def bulb_switcher_ii(number_of_bulbs, number_of_presses):
    if number_of_presses == 0:
        return 1
    if number_of_bulbs == 1:
        return 2
    if number_of_bulbs == 2:
        if number_of_presses == 1:
            return 3
        else:
            return 4
    if number_of_bulbs >= 3:
        if number_of_presses == 1:
            return 4
        elif number_of_presses == 2:
            return 7
        else:
            return 8

def bulb_switcher_ii_detailed(number_of_bulbs, number_of_presses):
    # Handle edge cases where the solution is trivial
    if number_of_presses == 0:
        return 1

    possible_states = set()

    # Iterate through all possible combinations of button presses
    for first_button in range(2 if number_of_presses > 0 else 1):
        for second_button in range(2 if number_of_presses > 0 else 1):
            for third_button in range(2 if number_of_presses > 0 else 1):
                for fourth_button in range(2 if number_of_presses > 0 else 1):
                    if sum([first_button, second_button, third_button, fourth_button]) <= number_of_presses:
                        # Simulate the button presses on the bulbs
                        bulbs = [1] * min(number_of_bulbs, 6) # Only the first 6 bulbs matter

                        if first_button:
                            for i in range(len(bulbs)):
                                bulbs[i] = 1 - bulbs[i]
                        if second_button:
                            for i in range(0, len(bulbs), 2):
                                bulbs[i] = 1 - bulbs[i]
                        if third_button:
                            for i in range(1, len(bulbs), 2):
                                bulbs[i] = 1 - bulbs[i]
                        if fourth_button:
                            for i in range(0, len(bulbs), 3):
                                bulbs[i] = 1 - bulbs[i]

                        # Convert the state of the bulbs to a tuple and add to set
                        possible_states.add(tuple(bulbs))

    # The number of unique states is the answer
    return len(possible_states)

Big(O) Analysis

Time Complexity

O(1) – The solution computes a fixed number of possible states based on the number of bulbs and button presses, up to a small constant number of bulbs and button presses. The algorithm does not iterate through all possible bulb configurations in relation to the input size. The operations performed are independent of the size of input n, with a constant amount of work for a fixed number of bulbs and button presses. Thus, the time complexity is O(1).

Space Complexity

O(1) – The provided steps indicate that the algorithm explicitly finds all possible states and stores them in a set. The number of bulbs considered is capped at 3, and the number of button presses is limited by the problem constraints. Therefore, the size of the set storing possible states is bounded by a constant, independent of the input number of bulbs 'n'. The algorithm does not utilize auxiliary data structures that scale with the input size, resulting in constant auxiliary space.

Edge Cases

Case	How to Handle
n = 0, presses = 0	Return 1 as there's only one state: all bulbs on.
n = 0, presses > 0	Return 1 as there are no bulbs to change.
n > 0, presses = 0	Return 1 as all bulbs are on initially.
n = 1, presses > 0	Return 2 as only two states are possible: on or off.
n = 2, presses = 1	Return 3, representing all on, first off, second off, and both off.
n = 3, presses = 1	Return 4, representing all possible combinations.
presses is large (e.g., > 10)	Optimize by using presses % 2, as the effect of presses beyond a certain number repeats.
n is very large and presses is also large	The number of bulbs beyond 3 doesn't matter much because patterns repeat, so treat n as min(n, 3) to optimize computations.

n = 0, presses = 0

How to Handle:

Return 1 as there's only one state: all bulbs on.

n = 0, presses > 0

How to Handle:

Return 1 as there are no bulbs to change.

n > 0, presses = 0

How to Handle:

Return 1 as all bulbs are on initially.

n = 1, presses > 0

How to Handle:

Return 2 as only two states are possible: on or off.

n = 2, presses = 1

How to Handle:

Return 3, representing all on, first off, second off, and both off.

n = 3, presses = 1

How to Handle:

Return 4, representing all possible combinations.

presses is large (e.g., > 10)

How to Handle:

Optimize by using presses % 2, as the effect of presses beyond a certain number repeats.

n is very large and presses is also large

How to Handle:

The number of bulbs beyond 3 doesn't matter much because patterns repeat, so treat n as min(n, 3) to optimize computations.