Use the same algorithm for the perl solution as for other languages.

1 files changed, 24 insertions, 8 deletions

diff --git a/challenge-097/abigail/perl/ch-2.pl b/challenge-097/abigail/perl/ch-2.pl
index b0d145069e..179745252b 100644
--- a/challenge-097/abigail/perl/ch-2.pl
+++ b/challenge-097/abigail/perl/ch-2.pl
@@ -25,18 +25,34 @@ GetOptions 's=i'  =>  \my $sections;
 die "-s option required" unless defined $sections && $sections > 0;
 
 #
-# To get the minimum number of flips, for each position take the
-# minimum of the number of 0s and the number of 1s, and sum over
-# all the positions
+# There is no need to split the input string into chunks,
+# we can leave it as is. The $i-th character of the $j-th section
+# is as position $j * $s_len + $i, where $s_len is the length
+# of a section.
+#
+# To calculate the minimum number of flips, for each position $i, we
+# calculate the number of 0s there are on the $i-th positions of
+# each section. Given the number of 0s, we can calculate the number
+# of 1s. The minimum number of flips required to get all the bits
+# on position equal is the minimum of the number of 0s and the number
+# of 1s.
+#
+# So, to calculate the minimum number of flips, for each postion,
+# we sum the minimum of the number of 0s and the number of 1s.
 #
 
 while (<>) {
-    my $sum = 0;
     chomp;
-    my @chunks = /.{$sections}/g;
-    for (my $i = 0; $i < length $chunks [0]; $i ++) {
-        my $zeros = grep {substr ($_, $i, 1) eq "0"} @chunks;
-        $sum += min $zeros, @chunks - $zeros;
+    my $s_len = length () / $sections;
+    my $sum = 0;
+    for my $i (0 .. $s_len - 1) {
+        my $zeros = 0;
+        for my $j (0 .. $sections - 1) {
+            my $index = $j * $s_len + $i;
+            $zeros ++ if substr ($_, $index, 1) eq "0";
+        }
+        my $ones = $sections - $zeros;
+        $sum += min ($zeros, $ones);
     }
     say $sum;
 }