A-M baby

Task/Find-the-missing-permutation/Python/find-the-missing-permutation-1.py

@@ -0,0 +1,8 @@
+from itertools import permutations
+
+given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
+           CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
+
+allPerms = [''.join(x) for x in permutations(given[0])]
+
+missing = list(set(allPerms) - set(given)) # ['DBAC']

Task/Find-the-missing-permutation/Python/find-the-missing-permutation-2.py

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+def missing_permutation(arr):
+  "Find the missing permutation in an array of N! - 1 permutations."
+
+  # We won't validate every precondition, but we do have some basic
+  # guards.
+  if len(arr) == 0: raise Exception("Need more data")
+  if len(arr) == 1:
+      return [arr[0][1] + arr[0][0]]
+
+  # Now we know that for each position in the string, elements should appear
+  # an even number of times (N-1 >= 2).  We can use a set to detect the element appearing
+  # an odd number of times.  Detect odd occurrences by toggling admission/expulsion
+  # to and from the set for each value encountered.  At the end of each pass one element
+  # will remain in the set.
+  missing_permutation = ''
+  for pos in range(len(arr[0])):
+      s = set()
+      for permutation in arr:
+          c = permutation[pos]
+          if c in s:
+            s.remove(c)
+          else:
+            s.add(c)
+      missing_permutation += list(s)[0]
+  return missing_permutation
+
+given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
+           CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
+
+print missing_permutation(given)

Task/Find-the-missing-permutation/Python/find-the-missing-permutation-3.py

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+>>> from collections import Counter
+>>> given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
+           CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
+>>> ''.join(Counter(x).most_common()[-1][0] for x in zip(*given))
+'DBAC'
+>>>

Task/Find-the-missing-permutation/Python/find-the-missing-permutation-4.py

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+>>> from pprint import pprint as pp
+>>> pp(list(zip(*given)), width=120)
+[('A', 'C', 'A', 'D', 'B', 'A', 'A', 'C', 'D', 'B', 'C', 'C', 'C', 'A', 'A', 'B', 'D', 'B', 'B', 'B', 'C', 'D', 'D'),
+ ('B', 'A', 'C', 'A', 'C', 'C', 'D', 'D', 'A', 'C', 'A', 'D', 'B', 'B', 'D', 'D', 'C', 'A', 'A', 'D', 'B', 'B', 'C'),
+ ('C', 'B', 'D', 'C', 'D', 'B', 'C', 'A', 'B', 'A', 'D', 'B', 'A', 'D', 'B', 'C', 'B', 'C', 'D', 'A', 'D', 'C', 'A'),
+ ('D', 'D', 'B', 'B', 'A', 'D', 'B', 'B', 'C', 'D', 'B', 'A', 'D', 'C', 'C', 'A', 'A', 'D', 'C', 'C', 'A', 'A', 'B')]
+>>> pp([Counter(x).most_common() for x in zip(*given)])
+[[('C', 6), ('B', 6), ('A', 6), ('D', 5)],
+ [('D', 6), ('C', 6), ('A', 6), ('B', 5)],
+ [('D', 6), ('C', 6), ('B', 6), ('A', 5)],
+ [('D', 6), ('B', 6), ('A', 6), ('C', 5)]]
+>>> pp([Counter(x).most_common()[-1] for x in zip(*given)])
+[('D', 5), ('B', 5), ('A', 5), ('C', 5)]
+>>> pp([Counter(x).most_common()[-1][0] for x in zip(*given)])
+['D', 'B', 'A', 'C']
+>>> ''.join([Counter(x).most_common()[-1][0] for x in zip(*given)])
+'DBAC'
+>>>