PWC 197
PWC 197
I did not attempt the challenges in Julia this time. I have solutions in Perl 5 and Raku.
Challenge 1 (Move Zero)
We are given a list of integers, say (3,0,1,0,5). We are asked to sort the list so that all zeros move to the end, and the remaining items are in their original relative order. In my example, the sorted list would be (3,1,5,0,0).
This is a one-liner in Perl 5 and Raku.
perl -wl -e 'my @o = sort { ($a == 0) ? 1 : ( ($b == 0) ? -1 : 0 ) } @ARGV; print "@o"' $@
The logic is via the block argument to sort. This takes two parameters $a and $b representing two elements in the list being compared. If the comparison returns -1, then $a precedes $b in the sorted list. If the comparison returns 1, then $b precedes $a in the sorted list. If the comparison returns 0, then the order doesn't matter. The sort pattern here follows this logic: If both $a and $b are non-zero, they can be in any relative order in the sorted list. Return 0. If $a is zero, then return 1. In this case, $b should precede $a. (It's fine even if $b is zero too). If $b is zero and $a is not, then return -1. In this case, $a should precede $b. This works in recent versions of Perl 5 (I'm not sure how recent), which keep the relative order of $a and $b the same as the original list if 0 is returned. With older versions of Perl 5, the relative order of $a and $b with 0 would be arbitrary, and there's no guarantee that the non-zero items would maintain their original relative order. Challenge 2 (Wiggle Sort)I have not quite solved the problem as specified by our host, and this time I also have not met the test examples. Let me first state the problem I have solved, and then I will state the difference from the specification. The problem I solved: We are given a list of integers say @list. We are asked to sort them so that $list[0] <= $list[1] >= $list[2] <= $list[3] ... The terms with even-numbered positions in the sorted list are less than or equal to the adjacent terms on both sides with odd-numbered positions (Positions here of course start with 0 as in Perl 5 or Raku). Our host's specification makes this a strict inequality: $list[0] < $list[1] > $list[2] ... There is no guarantee of a solution in that case (consider for example, a list with all members equal e.g., 1,1,1,1,1,1). Also (like with the weak inequalities), the solution if it exists is not guaranteed to be unique. This makes the test examples less useful, because there are several alternative answers that are as good as the one given. Coming back to the more general problem that I solved, I first attempted to do it via a block argument to sort as in Challenge 1. I could not achieve this, though I suspect that it should be possible. It is tricky because the comparison of $a and $b depends on their position in the list after it is sorted. The approach I ended up with was to first sort in descending order, $list[0] >= $list[1] >= $list[2] ... I then swap the elements for each pair of indices (0,1), (2,3), (4,5), ... giving $list[0] <= $list[1] >= $list[2] <= $list[3] ... exactly as we want. Here is my Perl 5 subroutine to do it (with a nested sub-subroutine to do the swapping of elements). The complete Perl 5 script is here. sub wiggle_sort { #-- subsub local *wiggle = sub { my ($n)=@_; ($n+1, $n); }; #-- wiggle_sort root sub my @list = @_; my @retval; @list = sort {$b <=> $a} @list; for (my $i = 0; ($i <= (scalar(@list)-2)); $i += 2) { push @retval, @list[&wiggle($i)]; } (@list % 2) && (push @retval, $list[-1]); @retval; } Raku is a direct cut-paste-edit translation, except that I moved the nested sub out to a separate subroutine (full Raku code here). sub wiggle_sort(@list) { my @retval; my @list_ = @list.sort({$^b <=> $^a}).flat; loop (my $i = 0; ($i <= (@list_.elems)-2); $i += 2) { push @retval, (@list_[&wiggle($i)]); } (@list_ % 2) && (push @retval, @list_[*-1]); @retval.List.flat; } #-- subsub sub wiggle (Int $n) { $n+1, $n; } |
Comments