From 2b52dd46834552ce48d7f2ee74433973bb1546ed Mon Sep 17 00:00:00 2001
From: Thomas Köhler <jean-luc@picard.franken.de>
Date: Mon, 9 Jan 2023 20:18:39 +0100
Subject: Add solution 199
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Signed-off-by: Thomas Köhler <jean-luc@picard.franken.de>
---
 challenge-199/jeanluc2020/blog-1.txt   |  1 +
 challenge-199/jeanluc2020/blog-2.txt   |  1 +
 challenge-199/jeanluc2020/perl/ch-1.pl | 44 ++++++++++++++++++++++++++++
 challenge-199/jeanluc2020/perl/ch-2.pl | 52 ++++++++++++++++++++++++++++++++++
 4 files changed, 98 insertions(+)
 create mode 100644 challenge-199/jeanluc2020/blog-1.txt
 create mode 100644 challenge-199/jeanluc2020/blog-2.txt
 create mode 100755 challenge-199/jeanluc2020/perl/ch-1.pl
 create mode 100755 challenge-199/jeanluc2020/perl/ch-2.pl

diff --git a/challenge-199/jeanluc2020/blog-1.txt b/challenge-199/jeanluc2020/blog-1.txt
new file mode 100644
index 0000000000..5773d6ea73
--- /dev/null
+++ b/challenge-199/jeanluc2020/blog-1.txt
@@ -0,0 +1 @@
+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-199-1.html
diff --git a/challenge-199/jeanluc2020/blog-2.txt b/challenge-199/jeanluc2020/blog-2.txt
new file mode 100644
index 0000000000..b526ba3dd6
--- /dev/null
+++ b/challenge-199/jeanluc2020/blog-2.txt
@@ -0,0 +1 @@
+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-199-2.html
diff --git a/challenge-199/jeanluc2020/perl/ch-1.pl b/challenge-199/jeanluc2020/perl/ch-1.pl
new file mode 100755
index 0000000000..5fe9804ff5
--- /dev/null
+++ b/challenge-199/jeanluc2020/perl/ch-1.pl
@@ -0,0 +1,44 @@
+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-199/#TASK1
+#
+# You are given a list of integers, @list.
+#
+# Write a script to find the total count of Good Pairs.
+#
+## A pair (i, j) is called good if list[i] == list[j] and i < j.
+#
+####################
+#
+# solution:
+#
+# The solution to this problem is quite simple. Walk from the first element to
+# the last with i, then walk from i to the last element with j, and take a note
+# of the pair if list[i] == list[j]. Nothing too wild.
+
+use strict;
+use warnings;
+
+# some examples
+my @lists = (
+   [1,2,3,1,1,3],
+   [1,2,3],
+   [1,1,1,1],
+   [],
+   [1,2,3,4,1,2,3,1,2,1]
+);
+
+foreach my $list (@lists) {
+   find_good_pairs(@$list);
+}
+
+
+sub find_good_pairs {
+   my @list = @_;
+   my $count = 0;
+   foreach my $i (0..$#list) {
+      foreach my $j ($i+1..$#list) {
+         $count++ if $list[$i] == $list[$j];
+      }
+   }
+   print "[" . join(",",@list) . "] returns $count\n";
+}
diff --git a/challenge-199/jeanluc2020/perl/ch-2.pl b/challenge-199/jeanluc2020/perl/ch-2.pl
new file mode 100755
index 0000000000..5bb32a12e7
--- /dev/null
+++ b/challenge-199/jeanluc2020/perl/ch-2.pl
@@ -0,0 +1,52 @@
+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-199/#TASK2
+#
+# You are given an array of integers, @array and three integers $x,$y,$z.
+#
+# Write a script to find out total Good Triplets in the given array.
+# 
+# A triplet array[i], array[j], array[k] is good if it satisfies the following conditions:
+# 
+## a) 0 <= i < j < k <= n (size of given array)
+## b) abs(array[i] - array[j]) <= x
+## c) abs(array[j] - array[k]) <= y
+## d) abs(array[i] - array[k]) <= z
+# 
+#####################################
+#
+# solution:
+#
+# The solution to this problem is a bit similar to the first one where we have
+# to walk the array with 2 variables; now we need three, and the check we have
+# to do for each iteration is a bit more complicated.
+#
+
+
+use strict;
+use warnings;
+
+# some examples
+
+my @examples = (
+   [[3,0,1,1,9,7], 7, 2, 3],
+   [[1,1,2,2,3],0,0,1]
+);
+
+foreach my $example (@examples) {
+   my ($list, $x, $y, $z) = @$example;
+   find_good_triplets($x, $y, $z, @$list);
+}
+
+sub find_good_triplets {
+   my ($x, $y, $z, @list) = @_;
+   my $count = 0;
+   foreach my $i (0..$#list) {
+      foreach my $j ($i+1..$#list) {
+         foreach my $k ($j+1..$#list) {
+            $count++ if abs($list[$i]-$list[$j]) <= $x and abs($list[$j]-$list[$k]) <= $y and abs($list[$i]-$list[$k]) <= $z;
+         }
+      }
+   }
+   print "[" . join(",",@list) . "] returns $count\n";
+}
+
-- 
cgit