X-Git-Url: https://projects.mako.cc/source/wordplay-cdsw-solutions/blobdiff_plain/4e09563052459f99272cfee99d44f8a9abce85cb..d57af8e41dd0a0a46fc411ad2488d029244921e1:/solution_6_advanced.py

diff --git a/solution_6_advanced.py b/solution_6_advanced.py
index 24927ad..5db3920 100644
--- a/solution_6_advanced.py
+++ b/solution_6_advanced.py
@@ -1,25 +1,21 @@
 import scrabble
 
-
-# Print the longest word where every digit is unique.
+# Print the longest word where every character is unique.
 # I used a Set for this. Don't worry: you didn't miss anything if you don't know
 # what a set is. We didn't teach it, but if you are reading this, you get a bonus!
 
-
 # A set is a container like a list or a dict, except that *each element can be stored only once*.
 # Think of it like the keys of a dict, except there isn't any value associated with each key.
 # I use Sets to count digits below. Feel free to look up the Set online and try it in the
 # interpreter.
 
-
 new_words = []
 for word in scrabble.wordlist:
     local_chars = {}
-    if len(word) == len(set(word)):  # Wait what!? See if you can figure out why this works.
-	new_words.append(word)
-
+    if len(word) == len(set(word)): # Wait what!? See if you can figure out why this works.
+        new_words.append(word)
 
-# Reuse my code for longest (in this case, the code to track all occurences, from the 
+# Reuse my code for longest word, (in this case, the code to track all occurences) from the
 # advanced solution.
 
 longest_so_far = []
@@ -27,8 +23,8 @@ length_of_longest_word = 0
 
 for word in new_words:
     if len(word) > length_of_longest_word:
-	length_of_longest_word = len(word)
-	longest_so_far = [word]
+        length_of_longest_word = len(word)
+        longest_so_far = [word]
     elif len(word) == length_of_longest_word:
         longest_so_far.append(word)