RogerBW's Blog

Perl Weekly Challenge 57: tree inversion and unique prefixes 05 May 2020

I’ve been doing the Perl Weekly Challenges. The latest involved inverting a binary tree and finding unique prefixes.

You are given a full binary tree of any height.

Write a script to invert the tree, by mirroring the children of every node, from left to right.

The input can be any sensible machine-readable binary tree format of your choosing, and the output should be the same format.

Well then, I'll just use the format I wrote for last week's problem… which also works on non-binary trees, which is handy.

my @tree=qw(1 .2 ..4 ..5 .3 ..6 ..7);

And the interpreter I wrote then too.

my %tree;
{
  my @parent;
  foreach my $index (0..$#tree) {
    $tree[$index] =~ /^(\.*)(\d+)/;
    my ($depth,$val)=(length($1),$2);
    $tree{$index}{value}=$val;
    if ($depth>0) {
      push @{$tree{$parent[$depth-1]}{children}},$index;
    }
    $parent[$depth]=$index;
  }
}

The reversal is trivial:

foreach my $k (keys %tree) {
  if (exists $tree{$k}{children}) {
    @{$tree{$k}{children}}=reverse @{$tree{$k}{children}};
  }
}

Then it's just a recursion to print the thing in some sort of good order.

tdump(\%tree,0,0);

sub tdump {
  my ($tree,$index,$depth)=@_;
  print '.' x $depth,$tree->{$index}{value},"\n";
  if (exists $tree->{$index}{children}) {
    foreach my $c (@{$tree->{$index}{children}}) {
      tdump($tree,$c,$depth+1);
    }
  }
}

BONUS

In addition to the above, you may wish to pretty-print your binary tree in a human readable text-based format similar to the following:

   1
/  \
 3    2
/ \  / \

7 6 5 4

This is actually distinctly more fiddly. And here I do give up generality and rely on the tree being both full and binary; there are no leaf nodes at less than the maximum depth. I also rely on each tree value being a single character.

my %d;
my @t=([0,0]);
while (@t) {
  my $i=shift @t;
  my ($index,$depth)=@{$i};
  push @{$d{$depth}},$index;
  if (exists $tree{$index}{children}) {
    push @t,map {[$_,$depth+1]} @{$tree{$index}{children}};
  }
}

my %c;
my @out;
my $d=max(keys %d);
while (1) {
  my @r;
  if (@out) {
    foreach my $i (0..$#{$d{$d}}) {
      my $si=$tree{$d{$d}[$i]}{children}[0] or die "need a full tree";
      $r[$i]=$c{$si}+2;
    }
  } else {
    my $noffsets=scalar @{$d{$d}};
    my $j=-2;
    while ($noffsets) {
      $j+=2;
      push @r,$j;
      $j+=4;
      push @r,$j;
      $noffsets-=2;
    }
  }
  my $str=' ' x (max @r);
  my $stru=' ' x ((max @r)-1);
  my $m=0;
  foreach my $i (0..$#{$d{$d}}) {
    substr($str,$r[$i],1)=$tree{$d{$d}[$i]}{value};
    $c{$d{$d}[$i]}=$r[$i];
    unless ($d==0) {
      if ($m%2==0) {
        substr($stru,$r[$i]+1,1)='/';
      } else {
        substr($stru,$r[$i]-1,1)='\\';
      }
      $m++;
    }
  }
  unshift @out,$str;
  if ($d>0) {
    unshift @out,$stru;
  }
  if ($d==0) {
    last;
  }
  $d--;
}

print map {"$_\n"} @out;

Part 2 was very much simpler:

Write a script to find the shortest unique prefix for each each word in the given list. The prefixes will not necessarily be of the same length.

Sample Input

[ "alphabet", "book", "carpet", "cadmium", "cadeau", "alpine" ]

Expected Output [ "alph", "b", "car", "cadm", "cade", "alpi" ]

my @input=qw(alphabet book carpet cadmium cadeau alpine);

OK, so let's make these into a hash. (In Perl6 I'd use a Set, but I had lots of other things to do so I didn't bother with P6 for either of this week's challenges.)

Then for each candidate length N I fill another hash keyed on the first N letters; any N-letter tuple that's unique, i.e. there's only one entry under it, can be dumped and the input word deleted for faster checking of the rest.

my @out;
my %input=map {$_ => 1} @input;
my $len=1;
while (%input) {
  my %k;
  map {push @{$k{substr($_,0,$len)}},$_} keys %input;
  foreach my $k (keys %k) {
    if (scalar keys @{$k{$k}}==1) {
      push @out,$k;
      delete $input{$k{$k}[0]};
    }
  }
  $len++;
}

I could order them by the original input, but didn't.

print map {"$_\n"} sort @out;

This will fail in the case where one word is the same as the start of another (e.g. "branch" and "branching"), but the problem itself breaks down in that case. A terminator character (i.e. something that appears in no word and can be added to the end of each) would make a solution possible.

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