Mathematical: Digital Root approach

data_structures/arrays/add_digits.rb

@@ -33,10 +33,10 @@ def add_digits(num)
   add_digits(sum)
 end
 
-# puts(add_digits(38))
+puts(add_digits(38))
 # # => 2
 
-# puts(add_digits(284))
+puts(add_digits(284))
 # # => 5
 
 #
@@ -58,4 +58,4 @@ puts(add_digits(38))
 # => 2
 
 puts(add_digits(284))
-# => 5
+# => 5

maths/add_digits.rb

@@ -0,0 +1,41 @@
+# Challenge name: Add Digits
+#
+# Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
+#
+# Example:
+#
+# Input: 38
+# Output: 2
+# Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
+#              Since 2 has only one digit, return it.
+#
+# Follow up:
+# Could you do it without any loop/recursion in O(1) runtime?
+# @param {Integer} num
+# @return {Integer}
+
+#
+# Approach: Mathematical: Digital Root
+# https://en.wikipedia.org/wiki/Digital_root
+#
+
+#
+# Complexity Analysis
+#
+# Time Complexity: O(1).
+# Space Complexity: O(1).
+
+def add_digits(num)
+  if num == 0
+    return 0
+  elsif num % 9 == 0
+    return 9
+  end
+  return num % 9
+end
+
+puts(add_digits(38))
+# # => 2
+
+puts(add_digits(284))
+# # => 5

maths/sum_of_digits.rb

@@ -1,19 +0,0 @@
-# Given a number, find sum of its digits.
-
-def digits_sum(n)
-  a = 0
-  sum = 0
-  until n.zero?
-    a = n % 10
-    sum += a
-    n /= 10
-  end
-  sum
-end
-
-puts 'Sum of digits of 3456 is ' + digits_sum(3456).to_s
-# Sum of digits of 3456 is 18
-puts 'Sum of digits of 1234 is ' + digits_sum(1234).to_s
-# Sum of digits of 1234 is 10
-puts 'Sum of digits of 9251321 is ' + digits_sum(9_251_321).to_s
-# Sum of digits of 9251321 is 23