feat: solve No.319

Author: BaffinLee

301-400/319. Bulb Switcher.md

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+# 319. Bulb Switcher
+
+- Difficulty: Medium.
+- Related Topics: Math, Brainteaser.
+- Similar Questions: Bulb Switcher II, Minimum Number of K Consecutive Bit Flips, Number of Times Binary String Is Prefix-Aligned, Find the Pivot Integer.
+
+## Problem
+
+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:
+
+![](https://assets.leetcode.com/uploads/2020/11/05/bulb.jpg)
+
+```
+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`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var bulbSwitch = function(n) {
+    return Math.floor(Math.sqrt(n));
+};
+```
+
+**Explain:**
+
+For the `k`th bulb, it is switched exactly a number of times that is a divisor of `k`. If `k` has an even number of divisors, then the final state of the `k`th light bulb is dark; if `k` has an odd number of divisors, then the final state of the `k`th light bulb is bright.
+
+
+For `k`, if it has divisor `x`, then there must be divisor `k/x`. Therefore, as long as at that time, divisors appear in "pairs". This means that only when `k` is a "perfect square number", it will have an odd number of divisors, otherwise it must have an even number of divisors.
+
+Therefore, we only need to find out the number of perfect square numbers in `1 ~ n`, and the answer is `sqrt(n)`
+
+**Complexity:**
+
+* Time complexity : O(1).
+* Space complexity : O(1).