diff --git a/DIRECTORY.md b/DIRECTORY.md
index c9eaeea..2700569 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -1,4 +1,7 @@
 
+## Bit Manipulation
+  * [Power Of Two](https://github.com/TheAlgorithms/Ruby/blob/master/bit_manipulation/power_of_two.rb)
+
 ## Ciphers
   * [Merkle Hellman Cryptosystem](https://github.com/TheAlgorithms/Ruby/blob/master/ciphers/merkle_hellman_cryptosystem.rb)
 
diff --git a/bit_manipulation/power_of_two.rb b/bit_manipulation/power_of_two.rb
new file mode 100644
index 0000000..21a7044
--- /dev/null
+++ b/bit_manipulation/power_of_two.rb
@@ -0,0 +1,68 @@
+# Power of 2
+#
+# Given an integer n, return true if it is a power of two. Otherwise, return false.
+#
+# An integer n is a power of two, if there exists an integer x such that n == 2^x.
+#
+# Example 1:
+# Input: n = 1
+# Output: true
+# Explanation: 2^0 = 1
+#
+# Example 2:
+# Input: n = 16
+# Output: true
+# Explanation: 2^4 = 16
+#
+# Example 3:
+# Input: n = 3
+# Output: false
+#
+# Example 4:
+# Input: n = 4
+# Output: true
+#
+# Example 5:
+# Input: n = 5
+# Output: false
+#
+# Constraints: -231 <= n <= 231 - 1
+# @param {Integer} n
+# @return {Boolean}
+#
+
+#
+# Approach 1: Bitwise operators: Turn off the Rightmost 1-bit
+#
+
+# Note that there are two ways of solving this problem via bitwise operations:
+# 1. How to get / isolate the rightmost 1-bit: x & (-x).
+# 2. How to turn off (= set to 0) the rightmost 1-bit: x & (x - 1).
+# In this approach, we're reproducing item 2.
+
+# Complexity Analysis
+#
+# Time complexity: O(1).
+# Space complexity: O(1).
+
+def is_power_of_two(n)
+  return false if n < 1
+
+  n & (n - 1) == 0
+end
+
+n = 1
+# Output: true
+puts is_power_of_two(n)
+n = 16
+# Output: true
+puts is_power_of_two(n)
+n = 3
+# Output: false
+puts is_power_of_two(n)
+n = 4
+# Output: true
+puts is_power_of_two(n)
+n = 5
+# Output: false
+puts is_power_of_two(n)
diff --git a/maths/power_of_two.rb b/maths/power_of_two.rb
index ffa77cc..f507aa5 100644
--- a/maths/power_of_two.rb
+++ b/maths/power_of_two.rb
@@ -33,15 +33,15 @@
 
 # Approach 1: Recursion
 #
-# Time Complexity: O(1)
+# Time Complexity: O(logn)
 #
 def is_power_of_two(n)
   if n == 1
-    return true
-  elsif n%2 == 0
-    is_power_of_two(n/2)
+    true
+  elsif n % 2 == 0
+    is_power_of_two(n / 2)
   else
-    return false
+    false
   end
 end
 
@@ -113,4 +113,4 @@ n = 4
 puts is_power_of_two(n)
 n = 5
 # Output: false
-puts is_power_of_two(n)
\ No newline at end of file
+puts is_power_of_two(n)