From 53f16ea5f040c744d301c658094b9573edff8f52 Mon Sep 17 00:00:00 2001
From: Joe Reich <reich.sj1@gmail.com>
Date: Sat, 19 Sep 2015 16:52:05 -0400
Subject: [PATCH] write fuzzy parser

---
 parser.rb | 57 ++++++++++++++++++++++++++++++++++++++++++++++++-------
 1 file changed, 50 insertions(+), 7 deletions(-)

diff --git a/parser.rb b/parser.rb
index 9396117..072b68e 100644
--- a/parser.rb
+++ b/parser.rb
@@ -1,13 +1,56 @@
 require 'pry'
 
-def kinda_like?(fuzzy_word, exact_word)
-  # implement with your code here
-  # breaking the method up into pieces is encouraged
+# 'f' for fuzzy, 'e' for exact word
+def kinda_like?(f,e)
+  # exactMatch(f,e)
+  f == e || middleJumbled?(f,e) || offByOne?(f,e)
 end
 
-#def another_private_method
-#end
+# check if you can turn the exact word into the fuzzy one by jumbling up all 
+# of the exact word's letters, while keeping the first and last as is
+def middleJumbled?(f,e)
+  len = e.length
+  if len >= 2
+    # turn "coffee" into /c[coffee]{3,}e/, for instance
+    # the {3,} prohibits cutting out a big chunk of the middle of the word
+    regex = Regexp.new "#{e[0]}[#{e}]{#{len-3},}#{e[len-1]}"
+    regex === f
+  else
+    false 
+  end
+end
+
+# checks whether a single substition, insertion, or deletion will turn the 
+# exact word into the fuzzy word
+def offByOne?(f,e)
+  if f.length == e.length
+    return singleSubstition?(f,e)
+  else
+    return singleInsertion?(f,e)
+  end
+end
 
-#def some_private_method
-#end
+# checks if a single letter-for-letter solution will do the trick
+def singleSubstition?(f,e)
+  differences = 0
+  for i in 0...f.length
+    if f[i] != e[i] then differences += 1 end
+    if differences > 1 then return false end
+  end
+  return true
+end
+
+# checks if a single insertion or deletion will do the trick
+def singleInsertion?(f,e)
+  longer = [f,e].max
+  shorter = [f,e].min
 
+  offset = 0
+  index = 0
+  # wherever the two words differ, increase the offset and try again
+  while index < shorter.length do
+    shorter[index] == longer[index+offset] ? index += 1 : offset += 1
+    if offset > 1 then return false end
+  end
+  return true
+end