diff --git a/DIRECTORY.md b/DIRECTORY.md
index 4eb3612..17bbb59 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -14,6 +14,7 @@
     * [Get Products Of All Other Elements](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/get_products_of_all_other_elements.rb)
     * [Remove Elements](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/remove_elements.rb)
     * [Shuffle Array](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/shuffle_array.rb)
+    * [Single Number](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/single_number.rb)
     * [Sort Squares Of An Array](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/sort_squares_of_an_array.rb)
     * [Two Sum](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/two_sum.rb)
     * [Two Sum Ii](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/two_sum_ii.rb)
diff --git a/data_structures/arrays/single_number.rb b/data_structures/arrays/single_number.rb
new file mode 100644
index 0000000..163b7b1
--- /dev/null
+++ b/data_structures/arrays/single_number.rb
@@ -0,0 +1,61 @@
+# Challenge name: Single Number
+#
+# Given a non-empty array of integers nums, every element appears twice
+# except for one. Find that single one.
+#
+# Follow up: Could you implement a solution with a linear runtime
+# complexity and without using extra memory?
+#
+# @param {Integer[]} nums
+# @return {Integer}
+
+# 
+# Approach 1: Hash map
+#
+# Time Complexity: O(n)
+#
+def single_number(nums)
+  result_hash = {}
+  nums.each do |num|
+    if result_hash[num]
+      result_hash[num] +=1 
+    else
+      result_hash[num] = 1
+    end
+  end
+
+  result_hash.each do |k, v|
+    return k if v == 1
+  end
+end
+
+nums = [2, 2, 1]
+puts(single_number(nums))
+# Output: 1
+nums = [4, 1, 2, 1, 2]
+puts(single_number(nums))
+# Output: 4
+nums = [1]
+puts(single_number(nums))
+# Output: 1
+
+#
+# Approach 2: Use Ruby .count()
+#
+# Time Complexity: O(n^2)
+#
+def single_number(nums)
+  nums.find do |num|
+    nums.count(num) == 1
+  end
+end
+
+nums = [2, 2, 1]
+puts(single_number(nums))
+# Output: 1
+nums = [4, 1, 2, 1, 2]
+puts(single_number(nums))
+# Output: 4
+nums = [1]
+puts(single_number(nums))
+# Output: 1
\ No newline at end of file