Solutions to the challenge 059.

challenge-059/wanderdoc/perl/ch-1.pl

@@ -0,0 +1,78 @@
+#!perl
+use strict;
+use warnings FATAL => qw(all);
+
+
+
+=prompt
+You are given a linked list and a value k. Write a script to partition the linked list such that all nodes less than k come before nodes greater than or equal to k. Make sure you preserve the original relative order of the nodes in each of the two partitions.
+For example:
+Linked List: 1 → 4 → 3 → 2 → 5 → 2
+k = 3
+Expected Output: 1 → 2 → 2 → 4 → 3 → 5.
+
+=cut
+
+
+my $K = shift || 3;
+my @input = map {(1 .. 10)[rand 10] } 1 .. 10; # (1, 4, 3, 2, 5, 2);
+print join(' -> ', @input), $/;
+
+
+# LL-Implementation.
+# https://www.slideshare.net/lembark/perly-linked-lists
+
+my $list = [[]];
+
+my $node = $list->[0];
+for my $val ( @input )
+{
+     @$node = ([], $val);
+     $node = $node->[0];
+}
+
+my $part_uK = [[]];
+my $node_uK = $part_uK->[0];
+
+my $part_oK = [[]];
+my $node_oK = $part_oK->[0];
+
+
+$node = $list->[0]; # Back to begin.
+
+while ($node)
+{
+     if ( $node->[1] and $node->[1] < $K )
+     {
+          @$node_uK = ([], $node->[1]);
+          $node_uK = $node_uK->[0]; 
+
+     }
+     elsif ( $node->[1] and $node->[1] >= $K )
+     {
+          @$node_oK = ([], $node->[1]);
+          $node_oK = $node_oK->[0];
+     }
+     $node = $node->[0];
+}
+
+@$node_uK = @{@$part_oK[0]}; # Now all the partitioned LL is in $part_uK.
+# $node_uK was already at the end of the list.
+
+# Free memory.
+undef $list; 
+undef $part_oK; 
+
+
+
+# Output.
+my $n_all = $part_uK->[0];
+while ( $n_all )
+{
+     print $n_all->[1] if $n_all->[1];
+
+     $n_all = $n_all->[0];
+     print ' -> ' if ($n_all and $n_all->[1]);
+
+}
+print $/;

challenge-059/wanderdoc/perl/ch-2.pl

@@ -0,0 +1,53 @@
+#!perl
+use strict;
+use warnings FATAL => qw(all);
+
+=prompt
+For this task, you will most likely need a function f(a,b) which returns the count of different bits of binary representation of a and b.
+For example, f(1,3) = 1, since:
+Binary representation of 1 = 01
+Binary representation of 3 = 11
+
+There is only 1 different bit. Therefore the subroutine should return 1. Note that if one number is longer than the other in binary, the most significant bits of the smaller number are padded (i.e., they are assumed to be zeroes).
+Script Output
+
+You script should accept n positive numbers. Your script should sum the result of f(a,b) for every pair of numbers given:
+For example, given 2, 3, 4, the output would be 6, since f(2,3) + f(2,4) + f(3,4) = 1 + 2 + 3 = 6
+
+=cut
+
+use List::Util qw(max);
+
+sub pairwise_difbits
+{
+     my ($n1, $n2) = @_;
+     my $max = max(map length(sprintf("%b", $_)), ($n1, $n2));
+     ($n1, $n2) =  map sprintf("%0${max}b", $_) , ($n1, $n2);
+
+     my $count = 0;
+     for my $i ( 0 .. $max - 1 )
+     {
+          $count++ if substr($n1, $i, 1) != substr($n2, $i, 1);
+     }
+     return $count;  
+}
+
+sub sumdif
+{
+     my @numbers = @_;
+     my $sum;
+
+
+     for my $i ( 0 .. $#numbers - 1 )
+     {
+          for my $j ( $i .. $#numbers )
+          {
+               next if $i == $j;
+               $sum += pairwise_difbits(@numbers[$i, $j]);
+          }
+     }
+
+     return $sum;
+}
+
+print sumdif ( 1, 2, 3, 4, 5 ), $/;