diff --git a/maths/fizz_buzz.rb b/maths/fizz_buzz.rb
index b35d5ee..133421c 100644
--- a/maths/fizz_buzz.rb
+++ b/maths/fizz_buzz.rb
@@ -54,3 +54,54 @@ fizz_buzz(n)
 #     "14",
 #     "FizzBuzz"
 # ]
+
+#
+# Approach 2: String Concatenation
+#
+
+# Algorithm
+#
+# Instead of checking for every combination of these conditions,
+# check for divisibility by given numbers i.e. 3, 5 as given in the
+# problem. If the number is divisible, concatenate the corresponding
+# string mapping Fizz or Buzz to the current answer string.
+#
+# For eg. If we are checking for the number 15, the steps would be:
+#
+# Condition 1: 15 % 3 == 0 , num_ans_str = "Fizz"
+# Condition 2: 15 % 5 == 0 , num_ans_str += "Buzz"
+# => num_ans_str = "FizzBuzz"
+#
+# So for FizzBuzz we just check for two conditions instead of three
+# conditions as in the first approach.
+#
+# Similarly, for FizzBuzzJazz now we would just have three
+# conditions to check for divisibility.
+
+# Complexity Analysis
+#
+# Time Complexity: O(N)
+# Space Complexity: O(1)
+
+# @param {Integer} n
+# @return {String[]}
+def fizz_buzz(n)
+  str = []
+
+  n.times do |i|
+    i += 1
+    num_str = ''
+
+    num_str += 'Fizz' if i % 3 == 0
+    num_str += 'Buzz' if i % 5 == 0
+
+    num_str = i.to_s if num_str == ''
+
+    str.push(num_str)
+  end
+
+  str
+end
+
+n = 15
+puts(fizz_buzz(n))