RogerBW's Blog

I’ve been doing the Weekly Challenges. The latest involved array filtering. (Note that this ends today.)

Task 1: Finishing Order

You are given two array of integers, @order and @friends.

The @order (size n) contains every integer from 1 to n exactly once. The @friends contains ID in sorted in strictly increasing order.

Write a script to return the array containing IDs in finishing order.

This is basically a filter: return the items in order that also appear in friends, in the order in which they appear in order. (So not quite a set intersection.)

Viewed this way it's quite easy, as in Crystal:

def finishingorder(order, friends)
  wanted = friends.to_set
  order.select{|x| wanted.includes?(x)}
end

(Some languages allow you to ask "does array X contain value Y" directly, but this feels ugly to me.)

Task 2: Trionic Array

You are given an array of integer.

Write a script to return true if the given array is Trionic otherwise return false.

An array is trionic if there exist indices 0 < p < q < n-1 such that arr[0…p] is strictly increasing, arr[p…q] is strictly decreasing, and arr[q…n-1] is strictly increasing.

Typst doesn't like early returns, and Scala can't even break out of a loop. Which is probably fine for the preferred functional style, and I dare say a Scala expert would say I shouldn't be writing this as an iterative state machine in the first place but rather looking at each sequential pair of values to see whether they're rising or falling. But I have states: 0 for not started yet, 1 for the first increase, 2 for the decrease, 3 for the second increase.

Languages with a skip (Rust, Crystal, Raku) make this easier.

Raku:

sub trionicarray(@a) {
    my $state = 0;
    my $prev = @a[0];
    for @a.skip(1) -> $n {
        my $dir = 0;
        if ($n > $prev) {
            $dir = 1;
        } elsif ($n < $prev) {
            $dir = -1;
        }
        if ($dir == 0) {
            return False;
        }
        given $state {
            when 0 {
                if ($dir == 1) {
                    $state = 1;
                } else {
                    return False;
                }
            }
            when 1 {
                if ($dir == -1) {
                    $state = 2;
                }
            }
            when 2 {
                if ($dir == 1) {
                    $state = 3;
                }
            }
            when 3 {
                if ($dir == -1) {
                    return False;
                }
            }
            default {
                return False;
            }
        }
        $prev = $n;
    }
    $state == 3;
}

Full code in all tagged languages is on codeberg.