Pip Stuart's submission for challenge-206.

4 files changed, 186 insertions, 0 deletions

diff --git a/challenge-206/pip/perl/ch-1.pl b/challenge-206/pip/perl/ch-1.pl
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+#!/usr/bin/perl
+# HTTPS://TheWeeklyChallenge.Org - Perl/Raku Weekly Challenge #206 - Pip Stuart
+# Task1: Shortest Time:  Submitted by: Mohammad S Anwar;  You are given a list of time points, at least 2, in the 24-hour clock format HH:MM.
+#   Write a script to find out the shortest time in minutes between any two time points.
+# Example1:
+#   In-put: @time = ("00:00", "23:55", "20:00"         )
+#   Output:  5        Since the difference between "00:00" and "23:55" is the shortest (5 minutes).
+# Example2:
+#   In-put: @time = ("01:01", "00:50", "00:57"         )
+#   Output:  4
+# Example3:
+#   In-put: @time = ("10:10", "09:30", "09:00", "09:55")
+#   Output: 15
+use strict;use warnings;use utf8;use v5.12;my $d8VS='N34LAnds';
+sub ShTm {my @time = @_;my $shtm = 0;my @tmnz = ();my %tdfz = ();my $sinc = '';
+  if     (   @time) {
+    for              (    0 ..($#time-1)) {
+      if (  $#time) {
+        for my $scnd (($_+1).. $#time   ) {
+          my @hhmm = split(/:/, $time[$_   ]);
+          $tmnz[$_   ] = $hhmm[0] * 60 + $hhmm[1];
+             @hhmm = split(/:/, $time[$scnd]);
+          $tmnz[$scnd] = $hhmm[0] * 60 + $hhmm[1];
+          if ($tmnz[$_   ] < $tmnz[$scnd]) {
+            $tdfz{ $tmnz[$scnd]             - $tmnz[$_   ] } = "\"$time[$_   ]\" and \"$time[$scnd]\"";
+            $tdfz{ $tmnz[$_   ] + (24 * 60) - $tmnz[$scnd] } = "\"$time[$scnd]\" and \"$time[$_   ]\"";
+          } else                           {
+            $tdfz{ $tmnz[$_   ]             - $tmnz[$scnd] } = "\"$time[$scnd]\" and \"$time[$_   ]\"";
+            $tdfz{ $tmnz[$scnd] + (24 * 60) - $tmnz[$_   ] } = "\"$time[$_   ]\" and \"$time[$scnd]\"";
+          } } } }
+    if   (  %tdfz ) {
+      my @stdf = sort { $a <=> $b } keys(%tdfz);
+      $shtm = $stdf[0]; # max shortest time should be is 720 minutes for just 2 times exactly 12 hours off from each other
+      $sinc = "Since the difference between $tdfz{$shtm} is the shortest ($shtm minutes).";
+    } }
+  printf("(%-34s) => %3d; %s\n", sprintf('"%s"', join('", "', @time)), $shtm, $sinc);
+  return($shtm);
+}
+if    (@ARGV) {
+  ShTm(@ARGV);
+} else {
+  ShTm("00:00", "23:55", "20:00"         ); # =>  5;
+  ShTm("01:01", "00:50", "00:57"         ); # =>  4;
+  ShTm("10:10", "09:30", "09:00", "09:55"); # => 15;
+}
diff --git a/challenge-206/pip/perl/ch-2.pl b/challenge-206/pip/perl/ch-2.pl
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index 0000000000..ffe7084a62
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+++ b/challenge-206/pip/perl/ch-2.pl
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+#!/usr/bin/perl
+# HTTPS://TheWeeklyChallenge.Org - Perl/Raku Weekly Challenge #206 - Pip Stuart
+# Task2: Array Pairings:  Submitted by: Mohammad S Anwar;  You are given an array of integers having even number of elements.
+#   Write a script to find the maximum sum of the minimum of each pairs.
+# Example1:
+#   In-put: @array = (1,2,3,4)
+#   Output:               4    Possible Pairings are as below:
+#     a) (1,2) and (3,4). So min(1,2) + min(3,4) => 1 + 3 => 4
+#     b) (1,3) and (2,4). So min(1,3) + min(2,4) => 1 + 2 => 3
+#     c) (1,4) and (2,3). So min(1,4) + min(2,3) => 2 + 1 => 3
+#   So the maximum sum is 4.
+# Example2:
+#   In-put: @array = (0,2,1,3)
+#   Output:               2    Possible Pairings are as below:
+#     a) (0,2) and (1,3). So min(0,2) + min(1,3) => 0 + 1 => 1
+#     b) (0,1) and (2,3). So min(0,1) + min(2,3) => 0 + 2 => 2
+#     c) (0,3) and (2,1). So min(0,3) + min(2,1) => 0 + 1 => 1
+#   So the maximum sum is 2.
+# Last date to submit the solution 23:59 (UK Time) Sunday 5th March 2023.
+use strict;use warnings;use utf8;use v5.12;my $d8VS='N34LHtJk';my @para=();my %sumz=();
+sub APrz {my @aray = @_; # recursively loop through all possible pairs
+  if     (!(@aray % 2)) {my $parz; # don't really need this pairz since TempARrAy below tracks removed pairs
+    for my $andx ( 1 .. $#aray ) {$parz = '';my @tara = @aray;
+      my $frst = shift (@tara); # pair up first value with each of the remaining values
+      my $scnd = splice(@tara, $andx - 1, 1);
+      push(@para,"$frst,$scnd"); # track pair progress
+      $parz   .= $para[-1];
+      if (@tara) { $parz .= APrz(@tara); } # recurse...
+      else       {my $psum = 0; # no pairs left so compute max sum of min of each pair
+        for (@para) {($frst,$scnd) = split(/,/,$_);
+          $psum += ($frst < $scnd) ? $frst : $scnd; }
+        $sumz{$psum} = join(' ', @para);
+#       say "@para => $psum;"; # this will print every set of pairs with their sum
+      } pop (@para);
+    } return($parz); } }
+sub MSMP {my @aray = @_;my $msmp = 0; # Max Sum of Min of Pairs
+  if     (!(@aray % 2)) {@para=();%sumz=(); # make sure there's an even number of elements to pair
+    APrz(@aray);
+    my @srts = sort { $a <=> $b } keys(%sumz);
+    $msmp = $srts[-1]; } # get Max Sum from last element of numerically sorted sumz
+  printf("(%-7s) => %d;\n", join(',', @aray), $msmp);
+  return($msmp); }
+if    (@ARGV) {
+  MSMP(@ARGV);
+} else {
+  MSMP(1,2,3,4); # => 4;
+  MSMP(0,2,1,3); # => 2;
+}
diff --git a/challenge-206/pip/raku/ch-1.raku b/challenge-206/pip/raku/ch-1.raku
new file mode 100644
index 0000000000..5b753a54c5
--- /dev/null
+++ b/challenge-206/pip/raku/ch-1.raku
@@ -0,0 +1,45 @@
+#!/usr/bin/env raku
+# HTTPS://TheWeeklyChallenge.Org - Perl/Raku Weekly Challenge #206 - Pip Stuart
+# Task1: Shortest Time:  Submitted by: Mohammad S Anwar;  You are given a list of time points, at least 2, in the 24-hour clock format HH:MM.
+#   Write a script to find out the shortest time in minutes between any two time points.
+# Example1:
+#   In-put: @time = ("00:00", "23:55", "20:00"         )
+#   Output:  5        Since the difference between "00:00" and "23:55" is the shortest (5 minutes).
+# Example2:
+#   In-put: @time = ("01:01", "00:50", "00:57"         )
+#   Output:  4
+# Example3:
+#   In-put: @time = ("10:10", "09:30", "09:00", "09:55")
+#   Output: 15
+use v6;my $d8VS='N35L6TKO';
+sub ShTm {my @time = @_;my $shtm = 0;my @tmnz = ();my %tdfz = ();my $sinc = '';
+  if     (   @time.elems) {
+    for     (    0 .. (@time.elems - 2))          {
+      if (   @time.elems - 1) {
+        for (($_+1).. (@time.elems - 1)) -> $scnd {
+          my @hhmm = split(':', @time[$_   ], :skip-empty);
+          @tmnz[$_   ] = @hhmm[0] * 60 + @hhmm[1];
+             @hhmm = split(':', @time[$scnd], :skip-empty);
+          @tmnz[$scnd] = @hhmm[0] * 60 + @hhmm[1];
+          if (@tmnz[$_   ] < @tmnz[$scnd]) {
+            %tdfz{ @tmnz[$scnd]             - @tmnz[$_   ] } = "\"@time[$_   ]\" and \"@time[$scnd]\"";
+            %tdfz{ @tmnz[$_   ] + (24 * 60) - @tmnz[$scnd] } = "\"@time[$scnd]\" and \"@time[$_   ]\"";
+          } else                           {
+            %tdfz{ @tmnz[$_   ]             - @tmnz[$scnd] } = "\"@time[$scnd]\" and \"@time[$_   ]\"";
+            %tdfz{ @tmnz[$scnd] + (24 * 60) - @tmnz[$_   ] } = "\"@time[$_   ]\" and \"@time[$scnd]\"";
+          } } } }
+    if   (  %tdfz ) {
+      my @stdf = sort +*, keys(%tdfz);
+      $shtm = @stdf[0]; # max shortest time should be is 720 minutes for just 2 times exactly 12 hours off from each other
+      $sinc = "Since the difference between %tdfz{$shtm} is the shortest ($shtm minutes).";
+    } }
+  printf("(%-34s) => %3d; %s\n", sprintf('"%s"', join('", "', @time)), $shtm, $sinc);
+  return($shtm);
+}
+if    (@*ARGS) {
+  ShTm(@*ARGS);
+} else {
+  ShTm("00:00", "23:55", "20:00"         ); # =>  5;
+  ShTm("01:01", "00:50", "00:57"         ); # =>  4;
+  ShTm("10:10", "09:30", "09:00", "09:55"); # => 15;
+}
diff --git a/challenge-206/pip/raku/ch-2.raku b/challenge-206/pip/raku/ch-2.raku
new file mode 100644
index 0000000000..0df8769050
--- /dev/null
+++ b/challenge-206/pip/raku/ch-2.raku
@@ -0,0 +1,48 @@
+#!/usr/bin/env raku
+# HTTPS://TheWeeklyChallenge.Org - Perl/Raku Weekly Challenge #206 - Pip Stuart
+# Task2: Array Pairings:  Submitted by: Mohammad S Anwar;  You are given an array of integers having even number of elements.
+#   Write a script to find the maximum sum of the minimum of each pairs.
+# Example1:
+#   In-put: @array = (1,2,3,4)
+#   Output:               4    Possible Pairings are as below:
+#     a) (1,2) and (3,4). So min(1,2) + min(3,4) => 1 + 3 => 4
+#     b) (1,3) and (2,4). So min(1,3) + min(2,4) => 1 + 2 => 3
+#     c) (1,4) and (2,3). So min(1,4) + min(2,3) => 2 + 1 => 3
+#   So the maximum sum is 4.
+# Example2:
+#   In-put: @array = (0,2,1,3)
+#   Output:               2    Possible Pairings are as below:
+#     a) (0,2) and (1,3). So min(0,2) + min(1,3) => 0 + 1 => 1
+#     b) (0,1) and (2,3). So min(0,1) + min(2,3) => 0 + 2 => 2
+#     c) (0,3) and (2,1). So min(0,3) + min(2,1) => 0 + 1 => 1
+#   So the maximum sum is 2.
+# Last date to submit the solution 23:59 (UK Time) Sunday 5th March 2023.
+use v6;my $d8VS='N35L6Who';my @para=();my %sumz=();
+sub APrz {my @aray = @_; # recursively loop through all possible pairs
+  if     (!(@aray % 2)) {my $parz; # don't really need this pairz since TempARrAy below tracks removed pairs
+    for  1 .. (@aray.elems-1) -> $andx {$parz = '';my @tara = @aray;
+      my $frst = shift (@tara); # pair up first value with each of the remaining values
+      my $scnd = splice(@tara, $andx - 1, 1);
+      push(@para,"$frst,$scnd"); # track pair progress
+      $parz   ~= @para[*-1];
+      if (@tara) { $parz ~= APrz(@tara); } # recurse...
+      else       {my $psum = 0; # no pairs left so compute max sum of min of each pair
+        for  @para  {($frst,$scnd) = split(',', $_, :skip-empty);
+          $psum += ($frst < $scnd) ?? $frst !! $scnd; }
+        %sumz{$psum} = join(' ', @para);
+#       say "@para[] => $psum;"; # this will print every set of pairs with their sum
+      }; pop (@para);
+    }; return($parz); } }
+sub MSMP {my @aray = @_;my $msmp = 0; # Max Sum of Min of Pairs
+  if     (!(@aray % 2)) {@para=();%sumz=(); # make sure there's an even number of elements to pair
+    APrz(@aray);
+    my @srts = sort +*, keys(%sumz);
+    $msmp = @srts[*-1]; } # get Max Sum from last element of numerically sorted sumz
+  printf("(%-7s) => %d;\n", join(',', @aray), $msmp);
+  return($msmp); }
+if    (@*ARGS) {
+  MSMP(@*ARGS);
+} else {
+  MSMP(1,2,3,4); # => 4;
+  MSMP(0,2,1,3); # => 2;
+}