Add solution 255.

Signed-off-by: Thomas Köhler <jean-luc@picard.franken.de>

2 files changed, 122 insertions, 0 deletions

diff --git a/challenge-255/jeanluc2020/python/ch-1.py b/challenge-255/jeanluc2020/python/ch-1.py
new file mode 100755
index 0000000000..728f307d69
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+++ b/challenge-255/jeanluc2020/python/ch-1.py
@@ -0,0 +1,62 @@
+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-255/#TASK1
+#
+# Task 1: Odd Character
+# =====================
+#
+# You are given two strings, $s and $t. The string $t is generated using the
+# shuffled characters of the string $s with an additional character.
+#
+# Write a script to find the additional character in the string $t..
+#
+## Example 1
+##
+## Input: $s = "Perl" $t = "Preel"
+## Output: "e"
+#
+## Example 2
+##
+## Input: $s = "Weekly" $t = "Weeakly"
+## Output: "a"
+#
+## Example 3
+##
+## Input: $s = "Box" $t = "Boxy"
+## Output: "y"
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Split the word into its character, store them in a hash table, then
+# count the result for each character in both the table for the original
+# word and for the new word.
+
+def odd_character(s: str, t: str) -> None:
+    print(f"Input: '{s}', '{t}'")
+    s_hash = {}
+    t_hash = {}
+    for char in list(s):
+        if char in s_hash:
+            s_hash[char] += 1
+        else:
+            s_hash[char] = 1
+    for char in list(t):
+        if char in t_hash:
+            t_hash[char] += 1
+        else:
+            t_hash[char] = 1
+    for found in t_hash.keys():
+        old = 0
+        if found in s_hash:
+            old = s_hash[found]
+        if t_hash[found] > old:
+            print(f"Output: {found}")
+            return
+
+
+odd_character("Perl", "Preel");
+odd_character("Weekly", "Weeakly");
+odd_character("Box", "Boxy");
diff --git a/challenge-255/jeanluc2020/python/ch-2.py b/challenge-255/jeanluc2020/python/ch-2.py
new file mode 100755
index 0000000000..ebc9f1edce
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+++ b/challenge-255/jeanluc2020/python/ch-2.py
@@ -0,0 +1,60 @@
+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-255/#TASK2
+#
+# Task 2: Most Frequent Word
+# ==========================
+#
+# You are given a paragraph $p and a banned word $w.
+#
+# Write a script to return the most frequent word that is not banned.
+#
+## Example 1
+##
+## Input: $p = "Joe hit a ball, the hit ball flew far after it was hit."
+##        $w = "hit"
+## Output: "ball"
+##
+## The banned word "hit" occurs 3 times.
+## The other word "ball" occurs 2 times.
+#
+## Example 2
+##
+## Input: $p = "Perl and Raku belong to the same family. Perl is the most
+## popular language in the weekly challenge."
+##        $w = "the"
+## Output: "Perl"
+##
+## The banned word "the" occurs 3 times.
+## The other word "Perl" occurs 2 times.
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Split the sentence into its words, then count all the words != $w
+
+import re
+from operator import itemgetter
+
+def most_frequent_word(p: str, w: str) -> None:
+    print(f"Input: '{p}', '{w}'")
+    found = {}
+    for word in re.split('\W+', p):
+        if word != w:
+            if word in found:
+                found[word] += 1
+            else:
+                found[word] = 1
+    most = []
+    for key in found.keys():
+        most.append( (key, found[key]) )
+    most.sort(key=itemgetter(1))
+    print(most)
+    print(f"Found the most frequent word to be '{most[-1][0]}'")
+
+most_frequent_word("Joe hit a ball, the hit ball flew far after it was hit.", "hit");
+most_frequent_word("Perl and Raku belong to the same family. Perl is the most" +
+   " popular language in the weekly challenge.", "the");
+