Challenge-059 solutions by saiftynet

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

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+#!/usr/env/perl
+# Task 1 Challenge 059 Solution by saiftynet
+
+# Linked List: PWC 059
+# k = 3# Expected Output: 1 → 2 → 2 → 4 → 3 → 5.
+# given a "linked list" partition about a number, but retain
+# relative order of other numbers
+
+my @l=qw/1 4 3 2 5 2/;
+
+print  join "->",partition(3);
+
+sub partition{
+
+# get the pivot and list
+  my ($k,@list)=@_;   
+
+# prepare anonymnous lists containing equal, after and before numbers
+  my @seq=([],[],[]); 
+
+# <=> return -1 if less (before), 0 if equal (pivot) and 1 of greater (after);
+# use this result as an index to push the numbers into one of these lists
+  foreach my $t (@l){
+    push @{$seq[$t<=>$k]},$t
+  };
+
+# return partitioned data
+  return @{$seq[-1]},@{$seq[0]},@{$seq[1]};
+
+}
+

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

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+#!/usr/env/perl
+# Task 2 Challenge 059 Solution by saiftynet
+# -1
+
+# Bitsum- sum of bits that are different.  This is effectively and xor
+# operation, followed by count of the bits that are 1.  One can convert 
+# the yiled of the xor operation into a binar string (e.g. using
+# sprintf "%b"), and counting the ones.  Getting the list context yield
+# of a match operation, the reading that in scalar context gives us what
+# we want.  A for loop that gets all possible pairs gives us a bit sum 
+# of a list of numbers
+
+print bitsum(2,3,4);
+sub bitsum{
+  my @list=@_;
+  my $sum=0;                         # accumulator
+  foreach my $m (0..@list-2){        # usual 2 fors to get
+    foreach my $n ($m+1..$#list){    # all combinations           
+     $sum    +=                      # get scalar
+            ()=                      # cast intoa array
+     (sprintf "%b", $list[$m]^$list[$n])# covert to binary
+          =~m/1/g;                   # get matches
+    }
+  }
+  return $sum;
+}