diff --git a/data_structures/arrays/strings/anagram_checker.rb b/data_structures/arrays/strings/anagram_checker.rb
index bcd5e36..1918b07 100644
--- a/data_structures/arrays/strings/anagram_checker.rb
+++ b/data_structures/arrays/strings/anagram_checker.rb
@@ -15,7 +15,7 @@
 # @return {Boolean}
 
 #
-# Approach 1: Sort and Compare
+# Approach: Sort and Compare
 #
 # Complexity analysis:
 #
diff --git a/data_structures/hash_table/anagram_checker.rb b/data_structures/hash_table/anagram_checker.rb
new file mode 100644
index 0000000..43163d4
--- /dev/null
+++ b/data_structures/hash_table/anagram_checker.rb
@@ -0,0 +1,55 @@
+# Challenge name: Is anagram
+#
+# Given two strings s and t , write a function to determine
+# if t is an anagram of s.
+#
+# Note:
+# You may assume the string contains only lowercase alphabets.
+#
+# Follow up:
+# What if the inputs contain unicode characters?
+# How would you adapt your solution to such case?
+#
+# @param {String} s
+# @param {String} t
+# @return {Boolean}
+
+#
+# Approach: Hash table
+#
+# Complexity analysis:
+#
+# Time complexity: O(n). Time complexity is O(n) since accessing the counter table is a constant time operation.
+# Space complexity: O(1). Although we do use extra space, the space complexity is O(1) because the table's size stays constant no matter how large n is.
+#
+def is_anagram(s, t)
+  s_length = s.length
+  t_length = t.length
+  counter = Hash.new(0)
+
+  return false unless s_length == t_length
+
+  (0...s_length).each do |i|
+    counter[s[i]] += 1
+    counter[t[i]] -= 1
+  end
+
+  counter.each do |k, v|
+    return false unless v == 0
+  end
+
+  true
+end
+
+s = 'anagram'
+t = 'nagaram'
+puts(is_anagram(s, t))
+# => true
+s = 'rat'
+t = 'car'
+puts(is_anagram(s, t))
+# => false
+s = 'a'
+t = 'ab'
+puts(is_anagram(s, t))
+# => false