RogerBW's Blog

Perl Weekly Challenge 73: fun with minima 15 August 2020

I’ve been doing the Perl Weekly Challenges. The latest involved finding minima in subsets of lists.

1. Min Sliding Window

You are given an array of integers @A and sliding window size $S.

Write a script to create an array of min from each sliding window. Example Input: @A = (1, 5, 0, 2, 9, 3, 7, 6, 4, 8) and $S = 3 Output: (0, 0, 0, 2, 3, 3, 4, 4)

[(1 5 0) 2 9 3 7 6 4 8] = Min (0) [1 (5 0 2) 9 3 7 6 4 8] = Min (0)

(etc.)

So the first thing I wanted was a fast minimum-finding function:

use List::Util qw(min);

(More people should use List::Util; it's great.)

Then it's just a matter of array-slicing the input and storing the result. No point in using a ring buffer for something this small.

sub msw {
  my $a=shift;
  my $s=shift;
  my @out;
  foreach my $i (0..(scalar @{$a})-$s) {
    push @out,min(@{$a}[$i..$i+$s-1]);
  }
  return \@out;
}

Raku is the same except for the usual array fiddliness.

sub msw(@a,$s) {
  my @out;
  for (0..(@a.elems-$s)) -> $i {
    push @out,min(@a[$i..$i+$s-1]);
  }
  return @out.flat;
}

2. Smallest Neighbour

You are given an array of integers @A.

Write a script to create an array that represents the smaller element to the left of each corresponding index. If none found then use 0.

Er, what?

But the examples made this clearer: "For each element N, if any element to the left of N is smaller than N, return the smallest of them; otherwise, return 0".

I'd love to know why one might want to do this (as with mystery stories, I like some narrative with my puzzle).

Anyway. List::Util again, obviously. At each step through the list we update the minimum value of all-entries-to-the-left (unlike problem we never remove anything from consideration so the minimum can only get lower), then push either that or zero onto the output. We short-circuit the first step since it's always zero.

sub sn {
  my $a=shift;
  my @out=(0);
  my $wm;
  foreach my $i (1..$#{$a}) {
    if (!defined $wm) {
      $wm=$a->[$i-1];
    } else {
      $wm=min($wm,$a->[$i-1]);
    }
    if ($wm < $a->[$i]) {
      push @out,$wm;
    } else {
      push @out,0;
    }
  }
  return \@out;
}

And Raku is mostly identical.


  1. Posted by Peter C at 09:21am on 15 August 2020

    I am amazed that you are still persisting with these. The puzzle-setter clearly ran out of ideas some while back, and the descriptions are badly-written and/or English is not their first language.

  2. Posted by RogerBW at 09:52am on 15 August 2020

    There are some interesting strangenesses in them; it's a good way to wake up on a Monday morning; and if the problem's an easy one, I try to do it with some style.

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