LeetCode 319: Bulb Switcher Solution
Master LeetCode problem 319 (Bulb Switcher), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
319. Bulb Switcher
Problem 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.
:
Solution Code
class Solution {
public int bulbSwitch(int n) {
return (int) Math.sqrt(n);
}
}Try It Yourself
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