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Leetcode 672. Bulb Switcher II

Problem Statement

You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are turned on. You can perform exactly one of four operations any number of times:

  1. Toggle all the bulbs.
  2. Toggle every second bulb.
  3. Toggle every third bulb.
  4. Toggle the bulbs with positions 1 + 3k, for all integers k (turning on the bulbs in positions 1, 4, 7, …).

You will return the number of different states after performing any number of operations.

Clarifying Questions

To understand the problem fully, here are some clarifying questions:

  1. Q: What does “toggle” mean in the context of this problem?
    • A: Toggling a bulb means if it is currently on (1), turn it off (0), and if it is currently off, turn it on.
  2. Q: Are all bulbs initially turned on?
    • A: Yes, all bulbs start in the on position (1).
  3. Q: What inputs are provided?
    • A: Two integers, n (number of bulbs) and m (the number of operations that can be performed).

Strategy

Let’s break down how we can approach this problem:

  1. Understand Bulb States: Each bulb can be either on or off, and specific operations toggle groups of bulbs.
  2. Pattern Observation: Given the constraints, we can find a pattern rather than simulating all operations which would be costly. Analyzing small values of n and m can help us deduce a pattern.
  3. Configuration Combinations: Depending on the value of n and m, the number of unique configurations can be fairly small. This results from the overlaps and redundancies of multiple operations.

Special Cases:

  1. If n == 0, there are no bulbs and hence 0 states.
  2. If m == 0, there is only one state where all bulbs are on.
  3. For larger values of n and m, configurations become a combination of the toggling operations.

Code

Below is the Java implementation of the logic described:

public class Solution {
    public int flipLights(int n, int m) {
        // Reduce problem size based on observation
        // States only depend on the first 6 bulbs effectively due to repeatable pattern
        if (n > 6) n = 6;
                    
        // Special case handling based on observed pattern
        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;
        
        return 8;
    }
}

Time Complexity

The time complexity of this solution is O(1), which is constant time. This is due to straightforward conditional checks that only evaluate a static number of conditions without looping or recursion.

The approach relies heavily on observed patterns rather than simulating the operation for every possible bulb and operation configuration, which offers a highly optimized solution in constant time.

By breaking down the problem and focusing on small-scale insights, this solution effectively leverages pattern recognition to offer a practical and efficient solution.

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