Adds problem79's solution

data/p079_keylog.txt

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+319
+680
+180
+690
+129
+620
+762
+689
+762
+318
+368
+710
+720
+710
+629
+168
+160
+689
+716
+731
+736
+729
+316
+729
+729
+710
+769
+290
+719
+680
+318
+389
+162
+289
+162
+718
+729
+319
+790
+680
+890
+362
+319
+760
+316
+729
+380
+319
+728
+716

problem79.py

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+#!/usr/bin/env python
+# -*- coding: utf-8 -*-
+
+
+"""
+A common security method used for online banking is to ask the user for three random characters from a passcode.
+For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters;
+the expected reply would be: 317.
+
+The text file, keylog.txt, contains fifty successful login attempts.
+
+Given that the three characters are always asked for in order, analyse the file so as to determine
+the shortest possible secret passcode of unknown length.
+"""
+
+from collections import defaultdict
+
+
+def pop_empty(dictionary):
+    for k, v in dictionary.iteritems():
+        if not v:
+            del dictionary[k]
+            return k
+    else:
+        raise ValueError(u"No empty key")
+
+
+def remove_value(dictionary, value):
+    for value_set in dictionary.itervalues():
+        if value in value_set:
+            value_set.remove(value)
+
+
+if __name__ == '__main__':
+    digits_preceding = defaultdict(set)
+
+    with open('data/p079_keylog.txt') as f:
+        for line in map(str.strip, f):
+            for idx in xrange(len(line)):
+                c = line[idx]
+                digits_preceding[c].update(line[:idx])
+
+    solution = ''
+
+    while digits_preceding:
+        first = pop_empty(digits_preceding)
+        remove_value(digits_preceding, first)
+        solution += first
+
+    print solution