Merge pull request #10359 from rjt-pl/master

rjt's Week 276 solutions and blog

6 files changed, 124 insertions, 4 deletions

diff --git a/challenge-276/ryan-thompson/README.md b/challenge-276/ryan-thompson/README.md
index 7a4e388b49..588de6d8b1 100644
--- a/challenge-276/ryan-thompson/README.md
+++ b/challenge-276/ryan-thompson/README.md
@@ -1,15 +1,17 @@
 # Ryan Thompson
 
-## Week 259 Solutions
+## Week 276 Solutions
 
-### Task 1 › Banking Day Offset
+### Task 1 › Complete Day
 
  * [Perl](perl/ch-1.pl)
+ * [Python](python/ch-1.py)
 
-### Task 2 › Line Parser
+### Task 2 › Maximum Frequency
 
  * [Perl](perl/ch-2.pl)
+ * [Python](python/ch-2.py)
 
 ## Blog
 
- * [Bank Holidays and Line Parser](https://ry.ca/2024/03/pwc-259-bank-holidays-and-line-parser/)
+ * [Maximum Frequency and now my Day is Complete](https://ry.ca/2024/07/pwc-276-complete-day-and-maximum-frequency/)
diff --git a/challenge-276/ryan-thompson/blog.txt b/challenge-276/ryan-thompson/blog.txt
new file mode 100644
index 0000000000..8e89370e93
--- /dev/null
+++ b/challenge-276/ryan-thompson/blog.txt
@@ -0,0 +1 @@
+https://ry.ca/2024/07/pwc-276-complete-day-and-maximum-frequency/
diff --git a/challenge-276/ryan-thompson/perl/ch-1.pl b/challenge-276/ryan-thompson/perl/ch-1.pl
new file mode 100755
index 0000000000..2bfd2b651e
--- /dev/null
+++ b/challenge-276/ryan-thompson/perl/ch-1.pl
@@ -0,0 +1,35 @@
+#!/usr/bin/env perl
+#
+# ch-1.pl - Complete Day
+#
+# 2024 Ryan Thompson <rjt@cpan.org>
+
+use 5.010;
+use warnings;
+use strict;
+no warnings 'uninitialized';
+use List::Util qw< sum0 reduce>;
+
+# If any pair of numbers is a multiple of 24, return the count of
+# such matching pairs. Assumption: pairs are WITH replacement.
+sub complete_day {
+    my $count = 0;
+    while (my $m = shift) {
+        $count += sum0 map { ($m + $_) % 24 == 0 } @_
+    }
+    $count
+}
+
+sub complete_day_functional {
+    sum0 map { my $m = shift @$_; map { ($m + $_) % 24 == 0 } @$_ }
+         map { [ @_[$_ .. $#_] ] } 0..$#_
+}
+
+use Test::More;
+
+is complete_day(qw<12 12 30 24 24>), 2 => 'Example 1';
+is complete_day(qw<72 48 24 5>),     3 => 'Example 2';
+is complete_day(qw<12 18 24>),       0 => 'Example 3';
+is complete_day(),                   0 => 'Empty list';
+
+done_testing;
diff --git a/challenge-276/ryan-thompson/perl/ch-2.pl b/challenge-276/ryan-thompson/perl/ch-2.pl
new file mode 100755
index 0000000000..e854457625
--- /dev/null
+++ b/challenge-276/ryan-thompson/perl/ch-2.pl
@@ -0,0 +1,34 @@
+#!/usr/bin/env perl
+#
+# ch-2.pl - Maximum Frequency
+#
+# 2024 Ryan Thompson <rjt@cpan.org>
+
+use 5.010;
+use warnings;
+use strict;
+no warnings 'uninitialized';
+use List::Util qw< sum0 max >;
+
+# 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)
+sub max_freq {
+    my %freq; # Frequency table
+    $freq{$_}++ for @_;
+
+    my $max_freq = max values %freq; # Maximal frequency
+    
+    $max_freq * grep { $_ == $max_freq } values %freq;
+}
+
+use Test::More;
+
+is max_freq(qw< 1 2 2 4 1 5 >), 4       => 'Example 1';
+is max_freq(qw< 1 2 3 4 5 >), 5         => 'Example 2';
+is max_freq(qw< 1 2 2 4 6 1 5 6 >), 6   => 'Example 3';
+is max_freq(qw< a a b a b b c >), 6     => 'Non-numeric';
+is max_freq(3.1415926), 1               => 'Single';
+is max_freq(), 0                        => 'Empty';
+
+done_testing;
diff --git a/challenge-276/ryan-thompson/python/ch-1.py b/challenge-276/ryan-thompson/python/ch-1.py
new file mode 100755
index 0000000000..9a217285c1
--- /dev/null
+++ b/challenge-276/ryan-thompson/python/ch-1.py
@@ -0,0 +1,19 @@
+#!/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
new file mode 100755
index 0000000000..08aff757c1
--- /dev/null
+++ b/challenge-276/ryan-thompson/python/ch-2.py
@@ -0,0 +1,29 @@
+#!/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