LeetCode 319: Bulb Switcher
LeetCode 319 Solution Explanation
Explanation:
To solve this problem, we need to understand that each bulb's state will only change when its index is a divisor of the current round number. For example, bulb 6 will only change its state in rounds 1, 2, 3, and 6.
We can observe that bulbs that are toggled an odd number of times will remain on, while bulbs toggled an even number of times will be off. Only perfect square numbers have an odd number of divisors. Therefore, we need to count how many perfect squares are less than or equal to n to determine the number of bulbs that will remain on.
The time complexity of this approach is O(1) since we are just calculating the square root of n. The space complexity is O(1) as we are not using any extra space.
:
LeetCode 319 Solutions in Java, C++, Python
class Solution {
public int bulbSwitch(int n) {
return (int) Math.sqrt(n);
}
}Interactive Code Editor for LeetCode 319
Improve Your LeetCode 319 Solution
Use the editor below to refine the provided solution for LeetCode 319. Select a programming language and try the following:
- Add import statements if required.
- Optimize the code for better time or space complexity.
- Add test cases to validate edge cases and common scenarios.
- Handle error conditions or invalid inputs gracefully.
- Experiment with alternative approaches to deepen your understanding.
Click "Run Code" to execute your solution and view the output. If errors occur, check the line numbers and debug accordingly. Resize the editor by dragging its bottom edge.
Loading editor...