finish 1.5

Author: camilosalazar98

Python/chapter01/1.5 - OneAway/camilo_solution_1.5.py

@@ -0,0 +1,75 @@
+"""
+One Away: There are three types of edits that can be performed on strings:
+insert a character, remove a character, or replace a character. Given two
+strings, write a function to check if they are one edit (or zero edits) away.
+EXAMPLE
+pale, ple -> true
+pales, pale -> true
+pale, bale -> true
+pale, bake -> false
+Hints:#23, #97, #130
+"""
+
+def oneAway(str1,str2):
+    str1 = str1.lower()
+    str2 = str2.lower()
+    d = {}
+    for i in str1:
+        if i in d:
+            d[i] =+ 1
+        else:
+            d[i] = 1
+
+    for j in str2:
+        if j in d:
+            d[j] -= 1
+
+        else:
+            d[j] = 1
+    count = 0# count the number of times
+
+
+    """
+    My thought on how to do this:
+    So knowing the pale and ple
+    when we subtract it from the hash table we know there should be a value of 1
+    for removal of one letter
+
+    The insertation of one letter should only leave the key of one
+    pales and pale
+
+    The interseting stuff happens when you have a change of letter.
+    The letter gets counting when you place it in the hash table
+    pale bale.
+    We use count to count the number of time count can be one. You'll find
+    that replaceing a letter is going to make count two v's of one
+
+    replacing two letter will make v's count more onces
+
+    You'll see this when you print k and v
+    
+    """
+    for k,v in d.items():
+        #print(k,v)
+        if v == 1:
+            count += 1
+
+    if count <= 2 : #
+        return True
+
+    else:
+        return False
+
+
+
+
+
+
+
+
+
+print(oneAway('pale', 'ple'))#True
+print(oneAway('pales', 'pale'))#True
+print(oneAway('pale', 'bale'))#True
+
+print(oneAway('pale', 'bake'))#False