Andrew Shitov Weekly Review: Challenge - 268

Tuesday, May 14, 2024| Tags: Raku

Raku Solutions Weekly Review


The first task was to find the magic number that, when added to every item of the first array, converts the array in such a way that all the numbers that you get after the addition are found in the second array of the same length. Sounds messy, but it effectively means that if you sort both arrays, then the difference between their corresponding elements is constant.

Let’s take the first example:

Input: @x = (3, 7, 5)
       @y = (9, 5, 7)
Output: 2

The magic number is 2.
@x = (3, 7, 5)
   +  2  2  2
@y = (5, 9, 7)

The sorted arrays are (3, 5, 7) and (5, 7, 9), and you can see now that 5 - 3 = 2, 7 - 5 = 2, and 9 - 7 = 2 too.

This is the principle on which I based my own solution. The tricky part is whether you want to give the answer by only computing the difference between the first elements of the sorted arrays, or you want to double check and confirm that the other items keep the same distance.

for @tests -> (@a is copy, @b is copy) {
    @a .= sort;
    @b .= sort;

    my $diff = @b[0] - @a[0];
    if @a >>+>> $diff eqv @b {
        say $diff;
    else {
        say "No such number";

I only compute the difference between the two first elements, and then test if the difference is applicable for all elements in the arrays.

Squish all

Arne Sommer demonstrates a few unique Raku syntax features in his solution.

First, the sorted input arrays are merged to make pairs, for which you can find the differences:

my @pairs = @x Z @y;
my @diff ={ $_[0] - $_[1] });

To confirm that all the differences are equal, a reduction metaoperator [==] is used:

say ( [==] @diff ) ?? @diff[0].abs !! 'error';

The construct [==] @diff is True if all the elements in @diff are equal.

Look at the soluton by Laurent Rosenfeld where [==] is used too:

sub magic-nr (@x, @y) {
    my @in1 = @x.sort;
    my @in2 = @y.sort;
    my @gaps = map {@in1[$_] - @in2[$_]}, 0..@x.end;
    return Nil unless [==] @gaps;
    return @gaps[0].abs;

Bruce Gray also used Z, but in combination with -, which allows to make both merging and computing differences at the same time: @y.sort Z- @x.sort.

sub task1_Z_minus ( @x, @y --> Numeric ) {
    die if @x.elems != @y.elems;

    my @y-x = (@y.sort Z- @x.sort).squish;

    return +@y-x == 1 ?? @y-x[0] !! Nil;

Here, the input is additionally checked to make sure the arrays have the same length.

The squish trick in this solution is very handy to avoid comparing the differences element by element. Consider this example:

my @x = (3, 7, 5);
my @y = (9, 5, 7);

my @y-x = (@y.sort Z- @x.sort);
dd @y-x; # Array @y-x = [2, 2, 2]

dd @y-x; # Array @y-x = [2]

The first output tells us that the differences are Array @y-x = [2, 2, 2]. We see that they are the same, so the second dd prints Array @y-x = [2] — and here is an array with the only value.

Now, update the intput values of @y:

my @x = (3, 7, 5);
my @y = (19, 5, 7); # 19 instead of 9 here

And this breaks the harmony: Array @y-x = [2, 12]. Differences are different, so a failure.

Luca Ferrari is squishing the arrays explicitly to checkif there is only single value in the end:

my @diffs;
for 0 ..^ @sorted-left.elems -> $i {
    my $current = @sorted-left[ $i ] - @sorted-right[ $i ];
    @diffs.push: $current if ! @diffs.grep( $current );

@diffs[ 0 ].say and exit  if @diffs.elems == 1;

Mark Anderson reminds us about another useful way to concurrently handle array elements in Raku, namely, using the >>-<< kind of operator:

my @r = @y.sort >>-<< @x.sort;
return .head if .elems == 1 given @r.squish;

We get the by-element differences in the @r variable, which then is squished and checked if it hosts a single value.

Another way to ensure all the differences are equal is used in the solution by Robert Ransbottom:

my @candi = (@a.sort [Z-] @b.sort);
return (@candi.all == @candi[0])
        ?? @candi[0].Int.abs
        !! Int;

Here, @candi.all == @candi[0] gives the answer to the question.

A similar approach is found in Jaldhar H. Vyas’s solution:

my @diff = $x.words.sort Z- $y.words.sort;
say @diff.all ?? @diff[0].abs !! "no magic number";

A more traditional way is used in the solution by Athanasius:

my Int $magic = @y-sorted[ 0 ] - @x-sorted[ 0 ];

for 1 .. $x.end -> UInt $i
    return unless @y-sorted[ $i ] - @x-sorted[ $i ] == $magic;

Packy Anderson evaluates the difference between the smallest items of the two arrays and then tests it against the other elements, quitting the loop if the check fails:

sub magicNumber(@x, @y) {
  my @xS = @x.sort;
  my @yS = @y.sort;
  my $magic = @yS.shift - @xS.shift;
  while (@xS) {
    if (@yS.shift - @xS.shift != $magic) {
      return; # no magic number
  return $magic;

Finally, Ulrich Rieke counts how many correct differences are there in the array of differences @zipped and if that agrees with the size of the original data:

if ( @zipped.grep( {$_[1] - $_[0] == $comp } ).elems == @firstNumbers.elems ) {
   say $comp ;

The Min-Min solutions

A few participants optimize the solution using the assumption that the input data is correct, so to find the difference, you even don’t need to sort the arrays, but rather find the minimums and use them directly:

Asher Harvey-Smith:

sub magic-number(Int:D @x, Int:D @y --> Int:D) {
    @y.min - @x.min

Jan Krnavek:

sub magic-number (@x, @y) {
    @y.min - @x.min


sub magic-number(@x, @y --> Int:D) {
    abs(min(@x) - min(@y))

Roger Bell_West:

sub magicnumber(@a, @b) {
    return min(@b) - min(@a);

By the way, notice that the TIMTOWTDI principle still works in Raku.

Bruce Gray is also using min in another variant of the solution:

sub task1_concise ( @x, @y --> Numeric ) {
    return adds_to_same(@x, $_, @y) ?? $_ !! Nil given @y.min - @x.min;

To make sure the difference is constant along the whole data, eqv is used to compare the baggified arrays:

sub adds_to_same { (@^a X+ $^addend).Bag eqv @^b.Bag }

One more

Before wrapping up, let us look at the solution by Feng Chang:

unit sub MAIN(Str:D $x, Str:D $y);

my @x = $x.comb(/\d+/);
my @y = $y.comb(/\d+/);
put (@y.sum - @x.sum) / +@x;

Here, none of above-seen bits are used. Instead, all the numbers in the arrays are added up, and the difference between the elements is computed as the difference of the sums divided by the size of the array.


If you have any suggestions or ideas then please do share with us.

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