Merge branch 'master' of https://github.com/manwar/perlweeklychallenge-club

11 files changed, 434 insertions, 0 deletions

diff --git a/challenge-285/jaldhar-h-vyas/blog.txt b/challenge-285/jaldhar-h-vyas/blog.txt
new file mode 100644
index 0000000000..3a342f4330
--- /dev/null
+++ b/challenge-285/jaldhar-h-vyas/blog.txt
@@ -0,0 +1 @@
+https://www.braincells.com/perl/2024/09/perl_weekly_challenge_week_285.html
diff --git a/challenge-285/jaldhar-h-vyas/perl/ch-1.pl b/challenge-285/jaldhar-h-vyas/perl/ch-1.pl
new file mode 100755
index 0000000000..48f35569c3
--- /dev/null
+++ b/challenge-285/jaldhar-h-vyas/perl/ch-1.pl
@@ -0,0 +1,25 @@
+#!/usr/bin/perl
+use v5.38;
+
+sub setDifference($arr1, $arr2) {
+    my %difference = map { $_ => 0 } @{$arr1};
+
+    for my $elem (@{$arr2}) {
+        if (exists $difference{$elem}) {
+            $difference{$elem}++;
+        }
+    }
+
+    return sort grep { !$difference{$_} } keys %difference;
+}
+
+my @first;
+my @second;
+
+for my $arg (@ARGV) {
+    @_ = split /\s+/, $arg;
+    push @first, $_[0];
+    push @second, $_[1];
+}
+
+say join q{ }, setDifference(\@second, \@first);
\ No newline at end of file
diff --git a/challenge-285/jaldhar-h-vyas/perl/ch-2.pl b/challenge-285/jaldhar-h-vyas/perl/ch-2.pl
new file mode 100755
index 0000000000..f394ead681
--- /dev/null
+++ b/challenge-285/jaldhar-h-vyas/perl/ch-2.pl
@@ -0,0 +1,32 @@
+#!/usr/bin/perl
+use v5.38;
+no warnings qw/ recursion /;
+use constant PENNY => 1;
+use constant NICKEL => 5;
+use constant DIME => 10;
+use constant QUARTER => 25;
+use constant HALFDOLLAR => 50;
+
+sub changeCombinations($amount, $largestCoin){
+    state @coins=(PENNY, NICKEL, DIME, QUARTER, HALFDOLLAR);
+    if ($largestCoin > $amount) {
+        return 0;
+    }
+
+    if ($largestCoin == $amount) {
+        return 1;
+    }
+
+    my $total;
+
+    for my $coin (grep {$_ <= $largestCoin} @coins) {
+        $total += changeCombinations($amount - $largestCoin, $coin);
+    }
+
+    return $total;
+}
+
+my ($amount) = @ARGV;
+my $largestCoin = HALFDOLLAR;
+
+say changeCombinations($amount + $largestCoin, $largestCoin)
diff --git a/challenge-285/jaldhar-h-vyas/raku/ch-1.sh b/challenge-285/jaldhar-h-vyas/raku/ch-1.sh
new file mode 100755
index 0000000000..4e779662a6
--- /dev/null
+++ b/challenge-285/jaldhar-h-vyas/raku/ch-1.sh
@@ -0,0 +1,3 @@
+#!/bin/sh
+
+raku -e 'my @a=@*ARGS.words.pairup;(@a.map({$_.value})∖@a.map({$_.key})).keys.join.say' "$@"
\ No newline at end of file
diff --git a/challenge-285/jaldhar-h-vyas/raku/ch-2.raku b/challenge-285/jaldhar-h-vyas/raku/ch-2.raku
new file mode 100755
index 0000000000..6c53fe4b65
--- /dev/null
+++ b/challenge-285/jaldhar-h-vyas/raku/ch-2.raku
@@ -0,0 +1,33 @@
+#!/usr/bin/raku
+
+constant PENNY = 1;
+constant NICKEL = 5;
+constant DIME = 10;
+constant QUARTER = 25;
+constant HALFDOLLAR = 50;
+
+sub changeCombinations($amount, $largestCoin){
+    constant @coins = (HALFDOLLAR, QUARTER, DIME, NICKEL, PENNY);
+
+    if $largestCoin > $amount {
+        return 0;
+    }
+
+    if $largestCoin == $amount {
+        return 1;
+    }
+
+    my $total;
+
+    for @coins.grep({ $_ <= $largestCoin }) -> $coin {
+        $total += changeCombinations($amount - $largestCoin, $coin);
+    }
+
+    return $total;
+}
+
+sub MAIN(
+    Int $amount
+) {
+    say changeCombinations($amount + HALFDOLLAR, HALFDOLLAR);
+}
\ No newline at end of file
diff --git a/challenge-285/lubos-kolouch/perl/ch-1.pl b/challenge-285/lubos-kolouch/perl/ch-1.pl
new file mode 100644
index 0000000000..63d294742a
--- /dev/null
+++ b/challenge-285/lubos-kolouch/perl/ch-1.pl
@@ -0,0 +1,51 @@
+#!/usr/bin/perl
+use strict;
+use warnings;
+use Test::More tests => 2;
+
+=pod
+
+=head1 DESCRIPTION
+
+This script finds the destination with no further outgoing connection from a given list of routes.
+
+=head1 FUNCTIONS
+
+=head2 find_destination(\@routes)
+
+Given an array reference of routes, where each route is an array reference containing a source and a destination city, this function returns the city that is a destination but not a source in any route.
+
+=over 4
+
+=item * C<\@routes> - Reference to an array of routes (array references of source and destination cities).
+
+=back
+
+Returns the destination city with no outgoing connections.
+
+=cut
+
+sub find_destination {
+    my ($routes_ref) = @_;
+    my %sources;
+    my %destinations;
+
+    foreach my $route (@$routes_ref) {
+        my ($source, $destination) = @$route;
+        $sources{$source} = 1;
+        $destinations{$destination} = 1;
+    }
+
+    foreach my $city (keys %destinations) {
+        unless (exists $sources{$city}) {
+            return $city;
+        }
+    }
+    return undef;  # In case there is no such city
+}
+
+# Unit Tests
+is(find_destination([["B","C"], ["D","B"], ["C","A"]]), "A", 'Example 1');
+is(find_destination([["A","Z"]]), "Z", 'Example 2');
+
+done_testing();
diff --git a/challenge-285/lubos-kolouch/perl/ch-2.pl b/challenge-285/lubos-kolouch/perl/ch-2.pl
new file mode 100644
index 0000000000..4ca1821de2
--- /dev/null
+++ b/challenge-285/lubos-kolouch/perl/ch-2.pl
@@ -0,0 +1,57 @@
+#!/usr/bin/perl
+use strict;
+use warnings;
+use Test::More tests => 3;
+
+=pod
+
+=head1 DESCRIPTION
+
+This script computes the number of distinct ways to make change for a given amount in cents using US coins.
+
+Coins available:
+- Penny (P): 1 cent
+- Nickel (N): 5 cents
+- Dime (D): 10 cents
+- Quarter (Q): 25 cents
+- Half-dollar (HD): 50 cents
+
+Order of coin selection does not matter.
+
+=head1 FUNCTIONS
+
+=head2 making_change($amount)
+
+Computes the number of ways to make change for the given amount.
+
+=over 4
+
+=item * C<$amount> - The amount in cents (non-negative integer).
+
+=back
+
+Returns the number of distinct ways to make change.
+
+=cut
+
+sub making_change {
+    my ($amount) = @_;
+    my @coins = (1, 5, 10, 25, 50);
+    my @dp = (0) x ($amount + 1);
+    $dp[0] = 1;
+
+    foreach my $coin (@coins) {
+        for my $i ($coin .. $amount) {
+            $dp[$i] += $dp[$i - $coin];
+        }
+    }
+
+    return $dp[$amount];
+}
+
+# Unit Tests
+is(making_change(9),   2,   'Example 1');
+is(making_change(15),  6,   'Example 2');
+is(making_change(100), 292, 'Example 3');
+
+done_testing();
diff --git a/challenge-285/lubos-kolouch/python/ch-1.py b/challenge-285/lubos-kolouch/python/ch-1.py
new file mode 100644
index 0000000000..be24d53a49
--- /dev/null
+++ b/challenge-285/lubos-kolouch/python/ch-1.py
@@ -0,0 +1,52 @@
+from typing import List
+import unittest
+
+
+def find_destination(routes: List[List[str]]) -> str:
+    """
+    Finds the destination with no further outgoing connection from a given list of routes.
+
+    Each route is represented as a list containing a source and a destination city.
+
+    Args:
+        routes (List[List[str]]): A list of routes.
+
+    Returns:
+        str: The destination city with no outgoing connections.
+
+    """
+    sources = set()
+    destinations = set()
+
+    for route in routes:
+        source, destination = route
+        sources.add(source)
+        destinations.add(destination)
+
+    no_outgoing = destinations - sources
+
+    if no_outgoing:
+        # Assuming there is only one such city
+        return no_outgoing.pop()
+    else:
+        return None  # In case there is no such city
+
+
+# Unit Tests
+class TestNoConnection(unittest.TestCase):
+
+    def test_example1(self):
+        routes = [["B", "C"], ["D", "B"], ["C", "A"]]
+        self.assertEqual(find_destination(routes), "A", 'Example 1')
+
+    def test_example2(self):
+        routes = [["A", "Z"]]
+        self.assertEqual(find_destination(routes), "Z", 'Example 2')
+
+    def test_no_destination(self):
+        routes = [["A", "B"], ["B", "A"]]
+        self.assertIsNone(find_destination(routes), 'No unique destination')
+
+
+if __name__ == "__main__":
+    unittest.main()
diff --git a/challenge-285/lubos-kolouch/python/ch-2.py b/challenge-285/lubos-kolouch/python/ch-2.py
new file mode 100644
index 0000000000..60182732c2
--- /dev/null
+++ b/challenge-285/lubos-kolouch/python/ch-2.py
@@ -0,0 +1,56 @@
+from typing import List
+import unittest
+
+
+def making_change(amount: int) -> int:
+    """
+    Compute the number of ways to make change for the given amount in cents using US coins.
+
+    Coins available:
+    - Penny (P): 1 cent
+    - Nickel (N): 5 cents
+    - Dime (D): 10 cents
+    - Quarter (Q): 25 cents
+    - Half-dollar (HD): 50 cents
+
+    Order of coin selection does not matter.
+
+    Args:
+        amount (int): The amount in cents (non-negative integer).
+
+    Returns:
+        int: The number of distinct ways to make change.
+    """
+    coins = [1, 5, 10, 25, 50]
+    dp = [0] * (amount + 1)
+    dp[0] = 1  # There is one way to make 0 cents
+
+    for coin in coins:
+        for i in range(coin, amount + 1):
+            dp[i] += dp[i - coin]
+
+    return dp[amount]
+
+
+# Unit Tests
+class TestMakingChange(unittest.TestCase):
+
+    def test_example1(self):
+        self.assertEqual(making_change(9), 2, 'Example 1')
+
+    def test_example2(self):
+        self.assertEqual(making_change(15), 6, 'Example 2')
+
+    def test_example3(self):
+        self.assertEqual(making_change(100), 292, 'Example 3')
+
+    def test_zero_amount(self):
+        self.assertEqual(making_change(0), 1, 'Zero amount')
+
+    def test_negative_amount(self):
+        with self.assertRaises(IndexError):
+            making_change(-1)
+
+
+if __name__ == "__main__":
+    unittest.main()
diff --git a/challenge-285/santiago-leyva/perl/ch-1.pl b/challenge-285/santiago-leyva/perl/ch-1.pl
new file mode 100644
index 0000000000..3b1b1ce918
--- /dev/null
+++ b/challenge-285/santiago-leyva/perl/ch-1.pl
@@ -0,0 +1,58 @@
+=begin
+You are given a list of routes, @routes.
+
+Write a script to find the destination with no further outgoing connection.
+
+Example 1
+Input: @routes = (["B","C"], ["D","B"], ["C","A"])
+Output: "A"
+
+"D" -> "B" -> "C" -> "A".
+"B" -> "C" -> "A".
+"C" -> "A".
+"A".
+Example 2
+Input: @routes = (["A","Z"])
+Output: "Z"
+=cut
+
+use strict;
+use Data::Dumper;
+
+my @input  = ([["B","C"], ["D","B"], ["C","A"]],[["A","Z"]]);
+
+foreach(@input){
+  my $arr = $_;
+  my @A = @$arr;
+  my $dead_end = findDeadEnd(\@A);
+  print $dead_end."\n";
+}
+
+sub findDeadEnd {
+  my $array = shift;
+  my @route = @$array;
+  my @origin;
+  my @destination;
+  my @deadend;
+  for(0..scalar @route-1){
+    push @origin, $route[$_][0];
+  }
+  for(0..scalar @route-1){
+    push @destination, $route[$_][1];
+  }
+
+  foreach my $i (@destination){
+    if( !(grep( /^$i$/, @origin)) ){
+      push @deadend, $i;
+    }
+    
+  }
+
+  if(scalar @deadend > 1){
+    return "There are multiple routes with no outgoing connection";
+  }elsif(scalar @deadend == 0){
+    return "All routes have an outgoing connection";
+  }
+  return $deadend[0];
+  
+}
diff --git a/challenge-285/santiago-leyva/perl/ch-2.pl b/challenge-285/santiago-leyva/perl/ch-2.pl
new file mode 100644
index 0000000000..6ebc42be82
--- /dev/null
+++ b/challenge-285/santiago-leyva/perl/ch-2.pl
@@ -0,0 +1,66 @@
+=begin
+Compute the number of ways to make change for given amount in cents. By using the coins e.g. Penny, Nickel, Dime, Quarter and Half-dollar, in how many distinct ways can the total value equal to the given amount? Order of coin selection does not matter.
+
+A penny (P) is equal to 1 cent.
+A nickel (N) is equal to 5 cents.
+A dime (D) is equal to 10 cents.
+A quarter (Q) is equal to 25 cents.
+A half-dollar (HD) is equal to 50 cents.
+Example 1
+Input: $amount = 9
+Ouput: 2
+
+1: 9P
+2: N + 4P
+Example 2
+Input: $amount = 15
+Ouput: 6
+
+1: D + 5P
+2: D + N
+3: 3N
+4: 2N + 5P
+5: N + 10P
+6: 15P
+Example 3
+Input: $amount = 100
+Ouput: 292
+=cut
+
+use strict;
+use Data::Dumper;
+
+use Test::More tests => 3;
+
+is(makeChange(9),2,'Test 1');
+is(makeChange(15),6,'Test 2');
+is(makeChange(100),292,'Test 3');
+
+
+my @amount = (9,15,100);
+
+foreach(@amount){
+  my $_last_coin;
+  my $combination = makeChange($_,$_last_coin);
+  print "Number of Combinations for $_ is: $combination\n";
+}
+
+sub makeChange { 
+  my ($reamaining,$last_coin) = @_;
+  my $combinations = 0;
+  my @denomination = (1,5,10,25,50);
+  #print "$reamaining,$last_coin\n";
+  foreach my $coin (@denomination){
+    if($last_coin and $last_coin < $coin){
+      #print "$last_coin < $coin next\n";
+      last;
+    }elsif($coin == $reamaining){
+      $combinations += 1;
+    }elsif($coin < $reamaining){
+      $combinations += makeChange($reamaining-$coin,$coin);
+    }
+  }
+
+  return $combinations;
+
+}
\ No newline at end of file