Bulb Switcher LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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

Bulb Switcher Solutions

Time: O(1)
Space: O(1)

C++

class Solution {
 public:
  int bulbSwitch(int n) {
    // k-th bulb only be switched when k % i == 0.
    // so we can reiterate the problem:
    // to find # of number <= 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);
  }
};

Java

 class Solution {
  public int bulbSwitch(int n) {
    // k-th bulb only be switched when k % i == 0.
    // so we can reiterate the problem:
    // to find # of number <= 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);
  }
}

Python


class Solution:
  def bulbSwitch(self, n: int) -> int:
    return int(sqrt(n))

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