From d4525399d41dd57f21bc11a67ceb63bdb16bc146 Mon Sep 17 00:00:00 2001
From: Simon Green <mail@simon.green>
Date: Sun, 11 Feb 2024 17:05:04 +1100
Subject: Simon's solution to challenge 255

---
 challenge-255/sgreen/python/ch-1.py | 50 +++++++++++++++++++++++++++++++++++++
 challenge-255/sgreen/python/ch-2.py | 38 ++++++++++++++++++++++++++++
 challenge-255/sgreen/python/test.py | 20 +++++++++++++++
 3 files changed, 108 insertions(+)
 create mode 100755 challenge-255/sgreen/python/ch-1.py
 create mode 100755 challenge-255/sgreen/python/ch-2.py
 create mode 100755 challenge-255/sgreen/python/test.py

(limited to 'challenge-255/sgreen/python')

diff --git a/challenge-255/sgreen/python/ch-1.py b/challenge-255/sgreen/python/ch-1.py
new file mode 100755
index 0000000000..dc81356e84
--- /dev/null
+++ b/challenge-255/sgreen/python/ch-1.py
@@ -0,0 +1,50 @@
+#!/usr/bin/env python3
+
+import sys
+
+
+def frequency_count(s: str) -> dict:
+    """Counts the frequency of each letter
+
+    Args:
+        s (str): The string to use
+
+    Returns:
+        dict: Key is the letter, values is the number of occurrences.
+    """
+    freq = {}
+
+    for c in s:
+        freq[c] = freq.get(c, 0) + 1
+
+    return freq
+
+
+def odd_character(s: str, t: str) -> str:
+    """Finds extra character in second string
+
+    Args:
+        s (str): The first string
+        t(str): The second string
+
+    Returns:
+        str: The extra character
+    """
+    first_freq = frequency_count(s)
+    second_freq = frequency_count(t)
+
+    for c in second_freq:
+        if second_freq[c] > first_freq.get(c, 0):
+            # We have found the extra character
+            return c
+
+    return 'No extra characters!'
+
+
+def main():
+    result = odd_character(sys.argv[1], sys.argv[2])
+    print(result)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/challenge-255/sgreen/python/ch-2.py b/challenge-255/sgreen/python/ch-2.py
new file mode 100755
index 0000000000..1a04788c22
--- /dev/null
+++ b/challenge-255/sgreen/python/ch-2.py
@@ -0,0 +1,38 @@
+#!/usr/bin/env python3
+
+import re
+import sys
+
+
+def most_frequent_word(p: str, b: str) -> str:
+    """Find the most frequent word that isn't banned
+
+    Args:
+        p (str): The paragraph
+        b (str): The banned word
+
+    Returns:
+        str: The most frequent word
+    """
+
+    # Calculate the frequency of all the words
+    freq = {}
+    for word in re.findall(r"\w+(?:'\w+)?", p.lower()):
+        freq[word] = freq.get(word, 0) + 1
+
+    # Remove the bad word (if it exists)
+    freq.pop(b, None)
+
+    # Find the maximum occurrence of any word, and the word that has this
+    max_count = max(freq.values())
+    words = [k for k in freq if freq[k] == max_count]
+    return words[0]
+
+
+def main():
+    result = most_frequent_word(sys.argv[1], sys.argv[2])
+    print(result)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/challenge-255/sgreen/python/test.py b/challenge-255/sgreen/python/test.py
new file mode 100755
index 0000000000..af2c369e5f
--- /dev/null
+++ b/challenge-255/sgreen/python/test.py
@@ -0,0 +1,20 @@
+#!/usr/bin/env python3
+
+import unittest
+ch_1 = __import__('ch-1')
+ch_2 = __import__('ch-2')
+
+
+class TestClass(unittest.TestCase):
+    def test_ch_1(self):
+        self.assertEqual(ch_1.odd_character('Perl', 'Preel'), 'e')
+        self.assertEqual(ch_1.odd_character('Weekly', 'Weeakly'), 'a')
+        self.assertEqual(ch_1.odd_character('Box', 'Boxy'), 'y')
+
+    def test_ch_2(self):
+        self.assertEqual(ch_2.most_frequent_word('Joe hit a ball, the hit ball flew far after it was hit.', 'hit'), 'ball')
+        self.assertEqual(ch_2.most_frequent_word('Perl and Raku belong to the same family. Perl is the most popular language in the weekly challenge.', 'the'), 'perl')
+
+
+if __name__ == '__main__':
+    unittest.main()
-- 
cgit