Added ex44

Author: nickdotht

exercises/ex43.py

@@ -6,7 +6,6 @@
 words in them."""
 
 from collections import defaultdict
-from exec_time import exec_time
 
 def anagram_finder(file_name):
 
@@ -16,13 +15,13 @@ def anagram_finder(file_name):
 		for line in f:
 			words.append(line.rstrip())
 
+	# The point here is that every word that have the same letters and the
+	# same length will be the same when they are sorted
 	# For a good enough explanation of defaultdict, see this answer:
 	# http://goo.gl/NbwXOv or in your Python interpreter, do:
 	# >> help('collections.defaultdict')
 	# If you know how defaultdict works, you should understand this
 	# block of codes
-	# The point here is that every word that have the same letters and the
-	# same length will be the same when they are sorted
 	anadict = defaultdict(list)
 	for word in words:
 		key = ''.join(sorted(word))

exercises/ex44.py

@@ -0,0 +1,39 @@
+"""Your task in this exercise is as follows:
+
+Generate a string with N opening brackets ("[") and N closing brackets ("]"), 
+in some arbitrary order. Determine whether the generated string is balanced; 
+that is, whether it consists entirely of pairs of opening/closing brackets 
+(in that order), none of which mis-nest.
+Examples:
+
+   []        OK   ][        NOT OK
+   [][]      OK   ][][      NOT OK
+   [[][]]    OK   []][[]    NOT OK
+"""
+
+from random import randrange
+import re
+
+def brackets(n):
+	i, result, brackets = 0, '', '[]'
+
+	# Add brackets in an arbitrary order
+	while i < n*2:
+		result += brackets[randrange(len(brackets))]
+		i+=1
+
+	# Save our string of brackets
+	bracket_string = result
+
+	# Remove all found pairs of brackets using regex
+	while len(re.findall(r'\[\]', result)) > 0:
+		result = re.sub(r'\[\]', '', result)
+
+	# If after removing all pairs of brackets the string is still not empty
+	if len(result) > 0:
+		print bracket_string, 'NOT OK'
+	else:
+		print bracket_string, 'OK'
+
+#test
+brackets(3)