319. Bulb Switcher LeetCode Solution

Last updated: February 17, 2025

In this guide, you will get 319. Bulb Switcher LeetCode Solution with the best time and space complexity. The solution to Bulb Switcher problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Bulb Switcher solution in C++
Bulb Switcher solution in Java
Bulb Switcher solution in Python
Additional Resources

319. Bulb Switcher LeetCode Solution image

Problem Statement of Bulb Switcher

There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.
On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.
Return the number of bulbs that are on after n rounds.

Example 1:

Input: n = 3
Output: 1
Explanation: At first, the three bulbs are [off, off, off].
After the first round, the three bulbs are [on, on, on].
After the second round, the three bulbs are [on, off, on].
After the third round, the three bulbs are [on, off, off].
So you should return 1 because there is only one bulb is on.
Example 2:

Input: n = 0
Output: 0

Example 3:

Input: n = 1
Output: 1

Constraints:

0 <= n <= 109

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(1)

319. Bulb Switcher LeetCode Solution in C++

class Solution {
 public:
  int bulbSwitch(int n) {
    // The k-th bulb can only be switched when k % i == 0.
    // So, we can rephrase the problem:
    // To find number of numbers <= n that have odd factors.
    // Obviously, only square numbers have odd factor(s).
    // e.g. n = 10, only 1, 4, and 9 are square numbers that <= 10
    return sqrt(n);
  }
};
/* code provided by PROGIEZ */

319. Bulb Switcher LeetCode Solution in Java

class Solution {
  public int bulbSwitch(int n) {
    // The k-th bulb can only be switched when k % i == 0.
    // So, we can rephrase the problem:
    // To find number of numbers <= n that have odd factors.
    // Obviously, only square numbers have odd factor(s).
    // e.g. n = 10, only 1, 4, and 9 are square numbers that <= 10
    return (int) Math.sqrt(n);
  }
}
// code provided by PROGIEZ

319. Bulb Switcher LeetCode Solution in Python

class Solution:
  def bulbSwitch(self, n: int) -> int:
    # The k-th bulb can only be switched when k % i == 0.
    # So, we can rephrase the problem:
    # To find number of numbers <= n that have odd factors.
    # Obviously, only square numbers have odd factor(s).
    # e.g. n = 10, only 1, 4, and 9 are square numbers that <= 10
    return math.isqrt(n)
 # code by PROGIEZ

Additional Resources

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