Leetcode 672: Bulb Switcher II

Input: The input consists of two integers `n` (the number of bulbs) and `presses` (the number of presses to make).

Example: n = 3, presses = 1

Constraints:

• 1 <= n <= 1000

• 0 <= presses <= 1000

Output: Return the number of distinct possible configurations of the bulbs after performing exactly `presses` presses.

Example: 4

Constraints:

• The result should be a non-negative integer.

Goal: Find the number of distinct possible configurations after exactly `presses` presses.

Steps:

• 1. Identify the effect of each button on the bulbs' states.

• 2. Calculate the number of distinct configurations possible based on the combination of button presses.

• 3. Return the number of distinct configurations as the result.

Goal: The number of bulbs and presses must be within the provided bounds.

Steps:

• 1 <= n <= 1000

• 0 <= presses <= 1000

Assumptions:

• The problem assumes that all bulbs are initially on and that presses are made sequentially.

• Input: n = 1, presses = 1

• Explanation: With one bulb and one press, the bulb can either be off (by pressing button 1) or stay on (by pressing button 2).

• Input: n = 2, presses = 1

• Explanation: With two bulbs and one press, the three possible configurations are [off, off], [on, off], and [off, on].

Approach: The solution involves simulating the effect of pressing the buttons and calculating the number of unique configurations.

Observations:

• Each button press can affect different bulbs, and some presses may result in the same configuration.

• We need to calculate the number of unique bulb configurations after all presses.

Steps:

• 1. For each button, track how it affects the bulbs.

• 2. Use combinatorics or a dynamic programming approach to calculate the number of unique configurations based on `presses`.

• 3. Return the total number of distinct configurations.

Empty Inputs:

• There are no empty inputs in this problem.

Large Inputs:

• Ensure the solution can handle the maximum values of `n` and `presses` (1000 each).

Special Values:

• If `presses` is 0, the configuration remains unchanged, and the result is 1.

Constraints:

• Handle edge cases where `n` is very small or `presses` is 0.

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

1 : Function Definition

int flipLights(int n, int m) {

This is the function definition for `flipLights`, which calculates the number of distinct configurations of `n` lights after `m` flips.

2 : Check if m is 0

    if(m==0) return 1;

If `m` is 0, meaning no flips are performed, the function returns 1, representing the initial configuration of the lights.

3 : Check if n is 1

    if(n==1) return 2;

If there is only one light (`n == 1`), there are two possible configurations: on or off, so the function returns 2.

4 : Handle Case for n=2 and m=1

    if(n==2&&m==1) return 3;

If there are two lights (`n == 2`) and one flip (`m == 1`), there are three possible configurations: both off, one on and the other off, or both on, so the function returns 3.

5 : Handle Case for n=2

    if(n==2) return 4;

If there are two lights (`n == 2`), and no other conditions are met, the function returns 4, representing all possible configurations (both off, one on, both on, and the reverse order of the previous configuration).

6 : Handle Case for m=1

    if(m==1) return 4;

If there is only one flip (`m == 1`), and no other conditions are met, the function returns 4, representing the configurations after one flip.

7 : Handle Case for m=2

    if(m==2) return 7;

If there are exactly two flips (`m == 2`), the function returns 7, representing the distinct light configurations that can be achieved with two flips.

8 : Handle Case for m>=3

    if(m>=3) return 8;

If there are three or more flips (`m >= 3`), the function returns 8, representing all possible configurations of the lights.

9 : Return Default Case

    return 8;

If none of the above conditions are met, return 8 as the default case, representing all possible configurations.

Best Case: O(n * presses)

Average Case: O(n * presses)

Worst Case: O(n * presses)

Description: The time complexity is based on simulating each press, which can involve looping through `n` bulbs for each of the `presses`.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to storing the states of the bulbs.

Leetcode 672: Bulb Switcher II

Solution to LeetCode 672: Bulb Switcher II Problem

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