Solve InterviewCake permutation-palindrome problem

Write a predicate to test whether any permutation of an input string is a
palindrome.

1 files changed, 37 insertions, 0 deletions

diff --git a/scratch/deepmind/part_two/permutation-palindrome.py b/scratch/deepmind/part_two/permutation-palindrome.py
new file mode 100644
index 000000000000..730b4bfdc873
--- /dev/null
+++ b/scratch/deepmind/part_two/permutation-palindrome.py
@@ -0,0 +1,37 @@
+import unittest
+from collections import Counter
+
+
+def has_palindrome_permutation(xs):
+    vs = Counter(xs).values()
+    return len([v for v in vs if v % 2 == 1]) in {0, 1}
+
+
+# Tests
+class Test(unittest.TestCase):
+    def test_permutation_with_odd_number_of_chars(self):
+        result = has_palindrome_permutation('aabcbcd')
+        self.assertTrue(result)
+
+    def test_permutation_with_even_number_of_chars(self):
+        result = has_palindrome_permutation('aabccbdd')
+        self.assertTrue(result)
+
+    def test_no_permutation_with_odd_number_of_chars(self):
+        result = has_palindrome_permutation('aabcd')
+        self.assertFalse(result)
+
+    def test_no_permutation_with_even_number_of_chars(self):
+        result = has_palindrome_permutation('aabbcd')
+        self.assertFalse(result)
+
+    def test_empty_string(self):
+        result = has_palindrome_permutation('')
+        self.assertTrue(result)
+
+    def test_one_character_string(self):
+        result = has_palindrome_permutation('a')
+        self.assertTrue(result)
+
+
+unittest.main(verbosity=2)