re 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 i^{th} round, you toggle every i bulb. For the n^{th} 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 <= 10^{9}

Bulb Switcher LeetCode Solution in C++

int bulbSwitch(int n) {
int counts = 0;
for (int i=1; i*i<=n; ++i) {
++ counts;
}
return counts;
}

public class Solution {
public int bulbSwitch(int n) {
return (int)Math.sqrt(n);
}
}

Bulb Switcher LeetCode Solution Review:

In our experience, we suggest you solve this Bulb Switcher LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Bulb Switcher LeetCode Solution

I hope this Bulb Switcher LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.