From af7708edbfe25d47cc7614748948ff9fc48d922d Mon Sep 17 00:00:00 2001
From: evenix <richardsondx@gmail.com>
Date: Wed, 2 Aug 2017 23:08:21 -0400
Subject: [PATCH] Update Q1_05_One_Away.rb

You can use the length of the intersection of the two string converted to an array to determine if there's one edit away.
---
 Chap_1_Arrays_and_Strings/Q1_05_One_Away.rb | 46 +++++++--------------
 1 file changed, 16 insertions(+), 30 deletions(-)

diff --git a/Chap_1_Arrays_and_Strings/Q1_05_One_Away.rb b/Chap_1_Arrays_and_Strings/Q1_05_One_Away.rb
index 9044372..c092709 100644
--- a/Chap_1_Arrays_and_Strings/Q1_05_One_Away.rb
+++ b/Chap_1_Arrays_and_Strings/Q1_05_One_Away.rb
@@ -1,34 +1,20 @@
 def one_away(s1, s2)
-	if s1 == s2
-		return true
-	elsif (s1.length - s2.length).abs > 1
-		return false
-	end
+  s1_array = s1.split('')
+  s2_array = s2.split('')
 
-	string_1 = s1.length >= s2.length ? s1 : s2
-	string_2 = s1.length < s2.length ? s1 : s2
+  intersection_length = (s1_array & s2_array).length
 
-	chars_1 = string_1.split('')
-	chars_2 = string_2.split('')
-
-	index_1 = 0
-	index_2 = 0
-	edit_count = 0
-
-	while (index_1 < string_1.length)
-		if chars_1[index_1] != chars_2[index_2]
-			edit_count += 1
-			index_2 += 1 if string_1.length == string_2.length
-		else
-			index_2 += 1
-		end
-
-		index_1 += 1
-
-		if edit_count > 1
-			return false
-		end
-	end
-
-	true
+  if s1.length >= s2.length
+    if intersection_length == (s1.length - 1) 
+      true
+    else
+      false
+    end
+  elsif s1.length < s2.length
+    if intersection_length == (s2.length - 1)
+      true
+    else
+      false
+    end 
+  end
 end