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diff --git a/challenge-276/ryan-thompson/python/ch-1.py b/challenge-276/ryan-thompson/python/ch-1.py
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+#!/usr/bin/env python3
+#
+# ch-1.py - Complete Day
+#
+# 2024 Ryan Thompson <rjt@cpan.org>
+
+def complete_day(hours):
+    count = 0
+    for i, m in enumerate(hours):
+        for n in filter(lambda n: ((m + n) % 24 == 0), hours[i+1:]):
+            count += 1
+
+    return(count)
+
+# Examples
+print(complete_day([12, 12, 30, 24, 24]))   # 2
+print(complete_day([72, 48, 24, 5]))        # 3
+print(complete_day([12, 18, 24]))           # 0
+print(complete_day([]))                     # 0 
diff --git a/challenge-276/ryan-thompson/python/ch-2.py b/challenge-276/ryan-thompson/python/ch-2.py
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+++ b/challenge-276/ryan-thompson/python/ch-2.py
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+#!/usr/bin/env python3
+#
+# ch-2.py - Maximum Frequency
+#
+# 2024 Ryan Thompson <rjt@cpan.org>
+
+# First identify which number(s) appear most often in the list,
+# then return the count of all such numbers. Ex. 1, 2, 2, 4, 1, 5
+# should return 4, because the matching numbers are 1,1,2,2 (freq:2)
+def max_freq(ints):
+    # Annoying special case for empty list
+    if len(ints) == 0:
+        return(0)
+
+    # Build the frequency table (freq[n] = # of times n is in ints)
+    freq = {} 
+    for n in ints: freq[n] = freq.setdefault(n,0) + 1
+
+    max_freq = max(freq.values()) # Maximal frequency
+
+    return(sum(filter(lambda x: x == max_freq, freq.values())))
+
+# Examples
+print(max_freq([1, 2, 2, 4, 1, 5]))                  # 4
+print(max_freq([1, 2, 3, 4, 5]))                     # 5
+print(max_freq([1, 2, 2, 4, 6, 1, 5, 6]))            # 6
+print(max_freq(['a', 'a', 'b', 'a', 'b', 'b', 'c'])) # 6
+print(max_freq([3.1415926]))                         # 1
+print(max_freq([]))                                  # 0