Simon's solution to challenge 194

6 files changed, 141 insertions, 2 deletions

diff --git a/challenge-194/sgreen/README.md b/challenge-194/sgreen/README.md
index f9ff4adc2a..f3687cb8b9 100644
--- a/challenge-194/sgreen/README.md
+++ b/challenge-194/sgreen/README.md
@@ -1,3 +1,3 @@
-# The Weekly Challenge 193
+# The Weekly Challenge 194
 
-Blog [The odd binary string](https://dev.to/simongreennet/the-odd-binary-string-p64)
+Blog [Digital frequency](https://dev.to/simongreennet/digital-frequency-l56)
diff --git a/challenge-194/sgreen/blog.txt b/challenge-194/sgreen/blog.txt
new file mode 100644
index 0000000000..ca6832d6c1
--- /dev/null
+++ b/challenge-194/sgreen/blog.txt
@@ -0,0 +1 @@
+https://dev.to/simongreennet/digital-frequency-l56
\ No newline at end of file
diff --git a/challenge-194/sgreen/perl/ch-1.pl b/challenge-194/sgreen/perl/ch-1.pl
new file mode 100644
index 0000000000..24e6929741
--- /dev/null
+++ b/challenge-194/sgreen/perl/ch-1.pl
@@ -0,0 +1,30 @@
+#!/usr/bin/env perl
+
+use strict;
+use warnings;
+use feature 'say';
+use experimental 'signatures';
+
+sub main($time) {
+    my $value = '';
+    if (substr($time, 0, 1) eq '?') {
+        # 2 if the second hour value is <= 3 else 1
+        $value = substr(time, 1, 1) <= 3 ? 2 : 1;
+    }
+    elsif (substr(time, 1, 1) eq '?') {
+        # 3 if the first hour value is 2 else 9
+        $value = substr(time, 1, 1) == 2 ? 3 : 9;
+    }
+    elsif (substr(time, 3, 1) eq '?') {
+        # The maximum first minute value is always 5
+        $value = 5;
+    }
+    else {
+        # The maximum second minute value is always 9
+        $value = 9;
+    }
+
+    say $value;
+}
+
+main($ARGV[0]);
\ No newline at end of file
diff --git a/challenge-194/sgreen/perl/ch-2.pl b/challenge-194/sgreen/perl/ch-2.pl
new file mode 100644
index 0000000000..e507fe4963
--- /dev/null
+++ b/challenge-194/sgreen/perl/ch-2.pl
@@ -0,0 +1,46 @@
+#!/usr/bin/env perl
+
+use strict;
+use warnings;
+use feature 'say';
+use experimental 'signatures';
+
+sub main($s) {
+    # Calculate the frequency of each letter
+    my %letter_freq = ();
+    foreach my $letter (split //, $s) {
+        $letter_freq{$letter} += 1;
+    }
+
+    # Calculate the frequency of frequencies :)
+    my %freq = ();
+    foreach my $i (values(%letter_freq)) {
+        $freq{$i} += 1;
+    }
+
+    my $solution = 0;
+    # A solution is only possible if there are two different frequencies ...
+    if (scalar keys(%freq) == 2) {
+        my $min_freq = min(keys(%freq));
+        my $max_freq = max(keys(%freq));
+
+        # ... and the minimum frequency only occurs once,
+        if ($min_freq == 1 and $freq{1} == 1) {
+            $solution = 1;
+        }
+
+        # ... or the difference between them is 1, and the higher frequency
+        #  only occurs once.
+        elsif ($min_freq == $max_freq - 1 and $freq{$max_freq} == 1) {
+            $solution = 1;
+        }
+    }
+    elsif (scalar keys(%freq) == 1 && exists($freq{1})) {
+        # ... or the only thing we have is single letters
+        $solution = 1;
+    }
+
+    say $solution;
+}
+
+main($ARGV[0])
\ No newline at end of file
diff --git a/challenge-194/sgreen/python/ch-1.py b/challenge-194/sgreen/python/ch-1.py
new file mode 100755
index 0000000000..add4e38257
--- /dev/null
+++ b/challenge-194/sgreen/python/ch-1.py
@@ -0,0 +1,23 @@
+#!/usr/bin/env python3
+
+import sys
+
+def main(time):
+    value = None
+    if time[0] == '?':
+        # 2 if the second hour value is <= 3 else 1
+        value = 2 if int(time[1]) <= 3 else 1
+    elif time[1] == '?':
+        # 3 if the first hour value is 2 else 9
+        value = 3 if time[0] == '2' else 9
+    elif time[3] == '?':
+        # The maximum first minute value is always 5
+        value = 5
+    else:
+        # The maximum second minute value is always 9
+        value = 9
+
+    print(value)
+
+if __name__ == '__main__':
+    main(sys.argv[1])
\ No newline at end of file
diff --git a/challenge-194/sgreen/python/ch-2.py b/challenge-194/sgreen/python/ch-2.py
new file mode 100755
index 0000000000..dfa570c4cf
--- /dev/null
+++ b/challenge-194/sgreen/python/ch-2.py
@@ -0,0 +1,39 @@
+#!/usr/bin/env python3
+
+import sys
+
+
+def main(s):
+    # Calculate the frequency of each letter
+    letter_freq = {}
+    for letter in s:
+        letter_freq[letter] = letter_freq.get(letter, 0) + 1
+
+    # Calculate the frequency of frequencies :)
+    freq = {}
+    for i in letter_freq.values():
+        freq[i] = freq.get(i, 0) + 1
+
+    solution = 0
+    # A solution is only possible if there are two different frequencies ...
+    if len(freq) == 2:
+        min_freq = min(freq.keys())
+        max_freq = max(freq.keys())
+
+        # ... and the minimum frequency only occurs once,
+        if min_freq == 1 and freq[1] == 1:
+            solution = 1
+
+        # ... or the difference between them is 1, and the higher frequency
+        #  only occurs once.
+        elif min_freq == max_freq - 1 and freq[max_freq] == 1:
+            solution = 1
+    elif len(freq) == 1 and 1 in freq:
+        # ... or the only thing we have is single letters
+        solution = 1
+
+    print(solution)
+
+
+if __name__ == '__main__':
+    main(sys.argv[1])