grid47 Exploring patterns and algorithms
Oct 6, 2024
3 min read
Solution to LeetCode 319: Bulb Switcher Problem
There are n bulbs initially off. You toggle every ith bulb in the ith round, and at the end, you need to return how many bulbs remain on.
Problem
Approach
Steps
Complexity
Input: You are given a single integer n, representing the number of bulbs.
Example: n = 4
Constraints:
• 0 <= n <= 10^9
Output: Return the number of bulbs that are still on after performing n rounds.
Example: 2
Constraints:
• The solution should handle large values of n efficiently.
Goal: To determine how many bulbs remain on after n rounds of toggling.
Steps:
• For each round i (1 to n), toggle every i-th bulb.
• A bulb remains on if it is toggled an odd number of times.
Goal: The solution must efficiently handle up to 10^9 rounds.
Steps:
• The solution must avoid simulating all rounds for large n, focusing on an optimized approach.
Assumptions:
• The number of bulbs can be very large (up to 10^9), so a direct simulation approach is impractical.
• Input: n = 4
• Explanation: The bulbs at positions 1 and 4 remain on after all rounds.
• Input: n = 9
• Explanation: The bulbs at positions 1, 4, and 9 remain on after all rounds.
Approach: Instead of simulating each round, we recognize that bulbs that are toggled an odd number of times will remain on. A bulb is toggled an odd number of times if its position is a perfect square.
Observations:
• Each bulb is toggled an odd number of times if and only if its position is a perfect square.
• The problem boils down to counting the number of perfect squares less than or equal to n.
Steps:
• Count the perfect squares less than or equal to n. This is equivalent to finding the integer part of the square root of n.
Empty Inputs:
• If n is 0, return 0.
Large Inputs:
• For large values of n, the solution must efficiently compute the integer square root without iterating through all bulbs.
Special Values:
• When n is a perfect square, ensure the count includes that square.
Constraints:
• The solution must efficiently handle up to 10^9 using the mathematical property of perfect squares.
intbulbSwitch(int n) {
return sqrt(n);
}
1 : Function Declaration
intbulbSwitch(int n) {
The function `bulbSwitch` takes an integer `n` as input, representing the number of bulbs, and returns an integer indicating the number of bulbs that remain on after the toggling process.
2 : Mathematical Operation
returnsqrt(n);
The function returns the square root of `n`, which represents the number of bulbs that remain on. This is based on the fact that only bulbs in positions that are perfect squares will remain on after toggling.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: The time complexity is constant because we only need to compute the integer square root of n.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant as we only store a few variables.
