How do you solve Bulb Switcher on LeetCode?

Bulb Switcher (LeetCode #319) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Only bulbs at positions that are perfect squares remain on after all rounds.

What is the brute force solution for Bulb Switcher?

The brute force approach for Bulb Switcher is Brute Force, with O(n²) time complexity and O(n) space complexity. We can simulate the process of toggling each bulb for every round. This approach is straightforward but inefficient for large n, as it requires multiple iterations over the bulb states. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Bulb Switcher?

Bulb Switcher uses the following concepts: Math, Brainteaser. The recommended patterns to study are: Math, Brainteaser.

What are the interview tips for Bulb Switcher?

Always consider edge cases, such as n = 0 or n = 1. Think about the mathematical properties of the problem rather than just simulating it. Practice explaining your reasoning clearly, as it helps solidify your understanding.

What are common mistakes when solving Bulb Switcher?

Confusing the toggling process with the final count of bulbs on. Not recognizing the relationship between divisors and bulb states.

#319

Bulb Switcher

Medium

MathBrainteaserMathBrainteaser

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n)	O(1)

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Intuition

Time O(1)Space O(1)

Instead of simulating each toggle, we can observe that a bulb's final state depends on the number of times it is toggled. A bulb at position k is toggled for each divisor it has. Only bulbs with an odd number of divisors remain on, which happens for perfect squares.

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Algorithm

3 steps

1Step 1: The bulbs that remain on correspond to the perfect squares up to n.
2Step 2: The number of perfect squares less than or equal to n is given by the integer part of the square root of n.
3Step 3: Return the integer part of the square root of n.

solution.py4 lines

1import math
2
3def bulbSwitch(n):
4    return int(math.sqrt(n))

ℹ

Complexity note: The time complexity is O(1) because we are only calculating the square root of n, which is a constant time operation. The space complexity is O(1) since we are using a fixed amount of space.

1Only bulbs at positions that are perfect squares remain on after all rounds.
2The number of bulbs that remain on is equal to the largest integer k such that k² ≤ n.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.