haoel · Vally79 · Dec 28, 2015 · Dec 28, 2015
diff --git a/README.md b/README.md
@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|319|[Bulb Switcher](https://leetcode.com/problems/bulb-switcher/) | [C++](./algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp)|Medium|
 |315|[Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/) | [C++](./algorithms/cpp/countOfSmallerNumbersAfterSelf/countOfSmallerNumbersAfterSelf.cpp)|Hard|
 |307|[Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable/) | [C++](./algorithms/cpp/rangeSumQuery-Immutable/rangeSumQuery-Mutable/RangeSumQueryMutable.cpp)|Medium|
 |306|[Additive Number](https://leetcode.com/problems/additive-number/) | [C++](./algorithms/cpp/additiveNumber/AdditiveNumber.cpp)|Medium|

diff --git a/algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp b/algorithms/cpp/bulbSwitcher/bulbSwitcher.cpp
@@ -0,0 +1,44 @@
+// Source : https://leetcode.com/problems/bulb-switcher/
+// Author : Calinescu Valentin
+// Date   : 2015-12-28
+
+/*************************************************************************************** 
+ *
+ * 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 nth round, you only toggle the
+ * last bulb. Find how many bulbs are on after n rounds.
+ * 
+ * Example:
+ * 
+ * Given n = 3. 
+ * 
+ * At first, the three bulbs are [off, off, off].
+ * After first round, the three bulbs are [on, on, on].
+ * After second round, the three bulbs are [on, off, on].
+ * After third round, the three bulbs are [on, off, off]. 
+ * 
+ * So you should return 1, because there is only one bulb is on.
+ * 
+ ***************************************************************************************/
+ /* 
+ * Solution 1 - O(1)
+ * =========
+ *
+ * We notice that for every light bulb on position i there will be one toggle for every
+ * one of its divisors, given that you toggle all of the multiples of one number. The 
+ * total number of toggles is irrelevant, because there are only 2 possible positions(on, 
+ * off). We quickly find that 2 toggles cancel each other so given that the start position
+ * is always off a light bulb will be in if it has been toggled an odd number of times.
+ * The only integers with an odd number of divisors are perfect squares(because the square
+ * root only appears once, not like the other divisors that form pairs). The problem comes
+ * down to finding the number of perfect squares <= n. That number is the integer part of
+ * the square root of n.
+ * 
+ */
+class Solution {
+public:
+    int bulbSwitch(int n) {
+        return (int)sqrt(n);
+    }
+};