Merge pull request #8806 from jeanluc2020/jeanluc-237

Add solution 237

6 files changed, 291 insertions, 0 deletions

diff --git a/challenge-237/jeanluc2020/blog-1.txt b/challenge-237/jeanluc2020/blog-1.txt
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+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-237-1.html
diff --git a/challenge-237/jeanluc2020/blog-2.txt b/challenge-237/jeanluc2020/blog-2.txt
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+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-237-2.html
diff --git a/challenge-237/jeanluc2020/perl/ch-1.pl b/challenge-237/jeanluc2020/perl/ch-1.pl
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+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-237/#TASK1
+#
+# Task 1: Seize The Day
+# =====================
+#
+# Given a year, a month, a weekday of month, and a day of week (1 (Mon) .. 7 (Sun)), print the day.
+#
+## Example 1
+##
+## Input: Year = 2024, Month = 4, Weekday of month = 3, day of week = 2
+## Output: 16
+##
+## The 3rd Tue of Apr 2024 is the 16th
+#
+## Example 2
+##
+## Input: Year = 2025, Month = 10, Weekday of month = 2, day of week = 4
+## Output: 9
+##
+## The 2nd Thu of Oct 2025 is the 9th
+#
+## Example 3
+##
+## Input: Year = 2026, Month = 8, Weekday of month = 5, day of week = 3
+## Output: 0
+##
+## There isn't a 5th Wed in Aug 2026
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# This is a case for the DateTime module:
+# - First we calculate the day of week for the 1st of the given month by
+#   creating a DateTime object
+# - Then we calculate the first day of the month that matches our target
+#   day of week
+# - From that, we calculate the weekday of month's date by adding another
+#   7 for each week we're out from the first week
+# - If we went past the end of the month, we return 0, otherwise the
+#   calculated date
+
+use strict;
+use warnings;
+use DateTime;
+
+
+seize_the_day(2024, 4, 3, 2);
+seize_the_day(2025, 10, 2, 4);
+seize_the_day(2026, 8, 5, 3);
+
+sub seize_the_day {
+   my ($year, $month, $weekday_of_month, $day_of_week) = @_;
+   print "Input: Year = $year, month = $month, weekday of month = $weekday_of_month, day of week = $day_of_week\n";
+   if($weekday_of_month < 1 or $weekday_of_month > 5) {
+      print "Output: 0\n";
+      return;
+   }
+   my $dt = DateTime->new( year => $year, month => $month, day => 1 );
+   my $days_per_month = {
+      1 => 31, 2 => $dt->is_leap_year ? 29 : 28,
+      3 => 31, 4 => 30, 5 => 31, 6 => 30, 7 => 31,
+      8 => 31, 9 => 30, 10 => 31, 11 => 30, 12 => 31
+   };
+   my $dow_1st = $dt->day_of_week;
+   my $first_appearance_of_dow = $day_of_week >= $dow_1st ? ( $day_of_week - $dow_1st + 1 ) : ( $day_of_week - $dow_1st + 8);
+   my $nth_appearance_of_dow = $first_appearance_of_dow + ($weekday_of_month-1) * 7;
+   if($nth_appearance_of_dow > $days_per_month->{$month}) {
+      print "Output: 0\n";
+      return;
+   }
+   print "Output: $nth_appearance_of_dow\n";
+}
+
diff --git a/challenge-237/jeanluc2020/perl/ch-2.pl b/challenge-237/jeanluc2020/perl/ch-2.pl
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index 0000000000..788060c5a9
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+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-237/#TASK2
+#
+# Task 2: Maximise Greatness
+# ==========================
+#
+# You are given an array of integers.
+#
+# Write a script to permute the given array such that you get the maximum
+# possible greatness.
+#
+### To determine greatness, nums[i] < perm[i] where 0 <= i < nums.length
+#
+## Example 1
+##
+## Input: @nums = (1, 3, 5, 2, 1, 3, 1)
+## Output: 4
+##
+## One possible permutation: (2, 5, 1, 3, 3, 1, 1) which returns 4 greatness as below:
+## nums[0] < perm[0]
+## nums[1] < perm[1]
+## nums[3] < perm[3]
+## nums[4] < perm[4]
+#
+## Example 2
+##
+## Input: @ints = (1, 2, 3, 4)
+## Output: 3
+##
+## One possible permutation: (2, 3, 4, 1) which returns 3 greatness as below:
+## nums[0] < perm[0]
+## nums[1] < perm[1]
+## nums[2] < perm[2]
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Calculate all permutations and check the greatness for each permutation
+# Keep the maximum
+
+use strict;
+use warnings;
+use Algorithm::Permute;
+
+maximise_greatness(1, 3, 5, 2, 1, 3, 1);
+maximise_greatness(1, 2, 3, 4);
+
+sub maximise_greatness {
+   my @ints = @_;
+   print "Input: (" . join(", ", @ints) . ")\n";
+   my $max = 0;
+   my $p_iterator = Algorithm::Permute->new ( \@ints );
+   while(my @perm = $p_iterator->next) {
+      my $current = greatness( [@_], [@perm] );
+      $max = $current if $current > $max;
+   }
+   print "Output: $max\n";
+}
+
+sub greatness {
+   my ($nums, $perm) = @_;
+   my @ints = @$nums;
+   my $greatness = 0;
+   foreach my $i (0..$#ints) {
+      $greatness++ if $nums->[$i] < $perm->[$i];
+   }
+   return $greatness;
+}
+
diff --git a/challenge-237/jeanluc2020/python/ch-1.py b/challenge-237/jeanluc2020/python/ch-1.py
new file mode 100755
index 0000000000..c8d33b1687
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+++ b/challenge-237/jeanluc2020/python/ch-1.py
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-237/#TASK1
+#
+# Task 1: Seize The Day
+# =====================
+#
+# Given a year, a month, a weekday of month, and a day of week (1 (Mon) .. 7 (Sun)), print the day.
+#
+## Example 1
+##
+## Input: Year = 2024, Month = 4, Weekday of month = 3, day of week = 2
+## Output: 16
+##
+## The 3rd Tue of Apr 2024 is the 16th
+#
+## Example 2
+##
+## Input: Year = 2025, Month = 10, Weekday of month = 2, day of week = 4
+## Output: 9
+##
+## The 2nd Thu of Oct 2025 is the 9th
+#
+## Example 3
+##
+## Input: Year = 2026, Month = 8, Weekday of month = 5, day of week = 3
+## Output: 0
+##
+## There isn't a 5th Wed in Aug 2026
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# This is a case for the datetime module:
+# - First we calculate the day of week for the 1st of the given month by
+#   creating a DateTime object
+# - Then we calculate the first day of the month that matches our target
+#   day of week
+# - From that, we calculate the weekday of month's date by adding another
+#   7 for each week we're out from the first week
+# - If we went past the end of the month, we return 0, otherwise the
+#   calculated date
+
+from datetime import date
+import calendar
+
+def seize_the_day(year: int, month: int, weekday_of_month: int, day_of_week: int):
+    print(f"Input: Year = {year}, month = {month}, weekday of month = {weekday_of_month}, day of week = {day_of_week}")
+    if weekday_of_month < 1 or weekday_of_month > 5:
+        print("Output: 0")
+        return
+    dtformat = "{year:4d}-{month:02d}-{day:02d}"
+    dt = date.fromisoformat(dtformat.format(year = year, month = month, day = 1))
+    days_per_month = { 1:31, 2:28, 3:31, 4:30, 5:31, 6:30, 7:31, 8:31, 9:30, 10:31, 11:30, 12:31 }
+    if calendar.isleap(year):
+        days_per_month[2] = 29
+
+    dow_1st = dt.isoweekday()
+    first_appearance_of_dow = day_of_week - dow_1st +1
+    if day_of_week < dow_1st:
+        first_appearance_of_dow += 7
+    nth_appearance_of_dow = first_appearance_of_dow + (weekday_of_month-1) * 7
+    if nth_appearance_of_dow > days_per_month[month]:
+        print("Output: 0")
+        return
+    print(f"Output: {nth_appearance_of_dow}")
+
+
+seize_the_day(year = 2024, month = 4,  weekday_of_month = 3, day_of_week = 2);
+seize_the_day(year = 2025, month = 10, weekday_of_month = 2, day_of_week = 4);
+seize_the_day(year = 2026, month = 8,  weekday_of_month = 5, day_of_week = 3);
+
+
diff --git a/challenge-237/jeanluc2020/python/ch-2.py b/challenge-237/jeanluc2020/python/ch-2.py
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index 0000000000..0125a8b508
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-237/#TASK2
+#
+# Task 2: Maximise Greatness
+# ==========================
+#
+# You are given an array of integers.
+#
+# Write a script to permute the given array such that you get the maximum
+# possible greatness.
+#
+### To determine greatness, nums[i] < perm[i] where 0 <= i < nums.length
+#
+## Example 1
+##
+## Input: @nums = (1, 3, 5, 2, 1, 3, 1)
+## Output: 4
+##
+## One possible permutation: (2, 5, 1, 3, 3, 1, 1) which returns 4 greatness as below:
+## nums[0] < perm[0]
+## nums[1] < perm[1]
+## nums[3] < perm[3]
+## nums[4] < perm[4]
+#
+## Example 2
+##
+## Input: @ints = (1, 2, 3, 4)
+## Output: 3
+##
+## One possible permutation: (2, 3, 4, 1) which returns 3 greatness as below:
+## nums[0] < perm[0]
+## nums[1] < perm[1]
+## nums[2] < perm[2]
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Calculate all permutations and check the greatness for each permutation
+# Keep the maximum
+
+from itertools import permutations
+
+def greatness(nums: list, perm: list) -> int:
+    greatness = 0
+    indices = list(range(len(nums)))
+    for i in indices:
+        if nums[i] < perm[i]:
+            greatness+=1
+    return greatness
+
+def maximise_greatness(ints: list):
+    print("Input: (" + ', '.join(str(x) for x in ints) + ")")
+    max = 0
+    for perm in permutations(ints):
+        current = greatness(ints, perm)
+        if current > max:
+            max = current
+    print(f"Output: {max}")
+
+
+maximise_greatness([1, 3, 5, 2, 1, 3, 1])
+maximise_greatness([1, 2, 3, 4])