Merge pull request #9016 from jeanluc2020/jeanluc-242

Add solution 242

6 files changed, 275 insertions, 0 deletions

diff --git a/challenge-242/jeanluc2020/blog-1.txt b/challenge-242/jeanluc2020/blog-1.txt
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+++ b/challenge-242/jeanluc2020/blog-1.txt
@@ -0,0 +1 @@
+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-242-1.html
diff --git a/challenge-242/jeanluc2020/blog-2.txt b/challenge-242/jeanluc2020/blog-2.txt
new file mode 100644
index 0000000000..97547946a0
--- /dev/null
+++ b/challenge-242/jeanluc2020/blog-2.txt
@@ -0,0 +1 @@
+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-242-2.html
diff --git a/challenge-242/jeanluc2020/perl/ch-1.pl b/challenge-242/jeanluc2020/perl/ch-1.pl
new file mode 100755
index 0000000000..d3b7aaf045
--- /dev/null
+++ b/challenge-242/jeanluc2020/perl/ch-1.pl
@@ -0,0 +1,62 @@
+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-242/#TASK1
+#
+# Task 1: Missing Members
+# =======================
+#
+# You are given two arrays of integers.
+#
+# Write a script to find out the missing members in each other arrays.
+#
+## Example 1
+##
+## Input: @arr1 = (1, 2, 3)
+##        @arr2 = (2, 4, 6)
+## Output: ([1, 3], [4, 6])
+##
+## (1, 2, 3) has 2 members (1, 3) missing in the array (2, 4, 6).
+## (2, 4, 6) has 2 members (4, 6) missing in the array (1, 2, 3).
+#
+## Example 2
+##
+## Input: @arr1 = (1, 2, 3, 3)
+##        @arr2 = (1, 1, 2, 2)
+## Output: ([3])
+##
+## (1, 2, 3, 3) has 2 members (3, 3) missing in the array (1, 1, 2, 2). Since they are same, keep just one.
+## (1, 1, 2, 2) has 0 member missing in the array (1, 2, 3, 3).
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# For each element of each array, if it is not in the other array,
+# save it for the output.
+# Example 2 has a caveat though, by not returning an empty array in
+# case of no missing elements it is unclear in which of the arrays
+# the element was originally. It's better to return an empty array
+# here, so let's just do that instead.
+
+missing_members( [ 1, 2, 3 ], [ 2, 4, 6 ] );
+missing_members( [ 1, 2, 3, 3 ], [ 1, 1, 2, 2 ] );
+
+sub missing_members {
+   my ($arr1, $arr2) = @_;
+   print "Input: (" . join(", ", @$arr1) . "), (" . join(", ", @$arr2) . ")\n";
+   my (@res1, @res2, %keys1, %keys2);
+   map { $keys1{$_} = 1; } @$arr1;
+   map { $keys2{$_} = 1; } @$arr2;
+   my %seen = ();
+   foreach my $elem (@$arr1) {
+      push @res1, $elem unless $keys2{$elem} or $seen{$elem};
+      $seen{$elem} = 1;
+   }
+   %seen = ();
+   foreach my $elem (@$arr2) {
+      push @res2, $elem unless $keys1{$elem} or $seen{$elem};
+      $seen{$elem} = 1;
+   }
+   print "Output: ([" . join(", ", @res1) . "], [" . join(", ", @res2) . "])\n";
+}
diff --git a/challenge-242/jeanluc2020/perl/ch-2.pl b/challenge-242/jeanluc2020/perl/ch-2.pl
new file mode 100755
index 0000000000..ed69f9eab7
--- /dev/null
+++ b/challenge-242/jeanluc2020/perl/ch-2.pl
@@ -0,0 +1,74 @@
+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-242/#TASK2
+#
+# Task 2: Flip Matrix
+# ===================
+#
+# You are given n x n binary matrix.
+#
+# Write a script to flip the given matrix as below.
+#
+# 1 1 0
+# 0 1 1
+# 0 0 1
+#
+# a) Reverse each row
+#
+# 0 1 1
+# 1 1 0
+# 1 0 0
+#
+# b) Invert each member
+#
+# 1 0 0
+# 0 0 1
+# 0 1 1
+#
+#
+## Example 1
+##
+## Input: @matrix = ([1, 1, 0], [1, 0, 1], [0, 0, 0])
+## Output: ([1, 0, 0], [0, 1, 0], [1, 1, 1])
+#
+## Example 2
+##
+## Input: @matrix = ([1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0])
+## Output: ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 0, 1], [1, 0, 1, 0])
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Just reverse and invert. Nothing complicated.
+
+flip_matrix([1, 1, 0], [1, 0, 1], [0, 0, 0]);
+flip_matrix([1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0]);
+
+sub flip_matrix {
+   my @rows = @_;
+   my @result;
+   print "Input: (";
+   foreach my $row (@rows) {
+      print "[" . join(", ", @$row) . "],";
+   }
+   print ")\n";
+   foreach my $row (@rows) {
+      push @result, [ inverted( reverse @$row ) ];
+   }
+   print "Ouput: (";
+   foreach my $row (@result) {
+      print "[" . join(", ", @$row) . "],";
+   }
+   print ")\n";
+}
+
+sub inverted {
+   my @array = @_;
+   my @result = ();
+   foreach my $elem (@array) {
+      push @result, $elem == 0 ? 1 : 0;
+   }
+   return @result;
+}
diff --git a/challenge-242/jeanluc2020/python/ch-1.py b/challenge-242/jeanluc2020/python/ch-1.py
new file mode 100755
index 0000000000..f17e000f6f
--- /dev/null
+++ b/challenge-242/jeanluc2020/python/ch-1.py
@@ -0,0 +1,66 @@
+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-242/#TASK1
+#
+# Task 1: Missing Members
+# =======================
+#
+# You are given two arrays of integers.
+#
+# Write a script to find out the missing members in each other arrays.
+#
+## Example 1
+##
+## Input: @arr1 = (1, 2, 3)
+##        @arr2 = (2, 4, 6)
+## Output: ([1, 3], [4, 6])
+##
+## (1, 2, 3) has 2 members (1, 3) missing in the array (2, 4, 6).
+## (2, 4, 6) has 2 members (4, 6) missing in the array (1, 2, 3).
+#
+## Example 2
+##
+## Input: @arr1 = (1, 2, 3, 3)
+##        @arr2 = (1, 1, 2, 2)
+## Output: ([3])
+##
+## (1, 2, 3, 3) has 2 members (3, 3) missing in the array (1, 1, 2, 2). Since they are same, keep just one.
+## (1, 1, 2, 2) has 0 member missing in the array (1, 2, 3, 3).
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# For each element of each array, if it is not in the other array,
+# save it for the output.
+# Example 2 has a caveat though, by not returning an empty array in
+# case of no missing elements it is unclear in which of the arrays
+# the element was originally. It's better to return an empty array
+# here, so let's just do that instead.
+
+def missing_members(arr1: list, arr2: list):
+    seen: dict = {}
+    res1: list = []
+    res2: list = []
+    print("Input: ([", ", ".join(str(x) for x in arr1), "], [", ", ".join(str(x) for x in arr2), "])")
+    for elem in arr1:
+        if elem in seen:
+            next
+        else:
+            if elem not in arr2:
+                seen[elem] = 1
+                res1.append(elem)
+    seen = {}
+    for elem in arr2:
+        if elem in seen:
+            next
+        else:
+            if elem not in arr1:
+                seen[elem] = 1
+                res2.append(elem)
+    print("Output: ([", ", ".join(str(x) for x in res1), "], [", ", ".join(str(x) for x in res2), "])")
+
+missing_members( [ 1, 2, 3 ], [ 2, 4, 6 ] );
+missing_members( [ 1, 2, 3, 3 ], [ 1, 1, 2, 2 ] );
+
diff --git a/challenge-242/jeanluc2020/python/ch-2.py b/challenge-242/jeanluc2020/python/ch-2.py
new file mode 100755
index 0000000000..df3a533199
--- /dev/null
+++ b/challenge-242/jeanluc2020/python/ch-2.py
@@ -0,0 +1,71 @@
+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-242/#TASK2
+#
+# Task 2: Flip Matrix
+# ===================
+#
+# You are given n x n binary matrix.
+#
+# Write a script to flip the given matrix as below.
+#
+# 1 1 0
+# 0 1 1
+# 0 0 1
+#
+# a) Reverse each row
+#
+# 0 1 1
+# 1 1 0
+# 1 0 0
+#
+# b) Invert each member
+#
+# 1 0 0
+# 0 0 1
+# 0 1 1
+#
+#
+## Example 1
+##
+## Input: @matrix = ([1, 1, 0], [1, 0, 1], [0, 0, 0])
+## Output: ([1, 0, 0], [0, 1, 0], [1, 1, 1])
+#
+## Example 2
+##
+## Input: @matrix = ([1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0])
+## Output: ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 0, 1], [1, 0, 1, 0])
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Just reverse and invert. Nothing complicated.
+
+def invert_reverse(array: list) -> list:
+    result: list = []
+    for x in reversed(array):
+        if x == 0:
+            result.append(1)
+        else:
+            result.append(0)
+    return result
+
+def flip_matrix(rows: list):
+    print("Input: (", end='')
+    for row in rows:
+        print("[", ", ".join(str(x) for x in row), "],", end='', sep='')
+    print(")")
+    result: list = []
+    for row in rows:
+        result.append(invert_reverse(row))
+    print("Output: (", end='')
+    for row in result:
+        print("[", ", ".join(str(x) for x in row), "],", end='', sep='')
+    print(")")
+
+flip_matrix([[1, 1, 0], [1, 0, 1], [0, 0, 0]])
+flip_matrix([[1, 1, 0, 0], [1, 0, 0, 1], [0, 1, 1, 1], [1, 0, 1, 0]])
+
+