Task 1: Max Positive Negative

You are given an array of integers, @ints.

Write a script to return the maximum number of either positive or negative integers in the given array.

Example 1

Input: @ints = (-3, 1, 2, -1, 3, -2, 4)
Output: 4

Count of positive integers: 4
Count of negative integers: 3
Maximum of count of positive and negative integers: 4

Example 2

Input: @ints = (-1, -2, -3, 1)
Output: 3

Count of positive integers: 1
Count of negative integers: 3
Maximum of count of positive and negative integers: 3

Example 3

Input: @ints = (1,2)
Output: 2

Count of positive integers: 2
Count of negative integers: 0
Maximum of count of positive and negative integers: 2

Approach

Really, this is two loops over the array of integers. One to count positive ints, one to count negative ints. If I do a map of the array each time and return 1 for each int I want to count and 0 for each int I don’t want to count, I can do the counting with a sum operator.

Raku

Like last week, we use Raku’s Reduction Metaoperator with addition ([+]) for the summation, and the max routine on the Any class to pick the maximum.

sub maxPosNeg(@ints) {
  my $pos = [+] @ints.map({ $_ > 0 ?? 1 !! 0 });
  my $neg = [+] @ints.map({ $_ < 0 ?? 1 !! 0 });
  my $max = max $pos, $neg;
  return (
    $max,
    (
      "Count of positive integers: $pos",
      "Count of negative integers: $neg",
      "Maximum of count of positive and " ~
      "negative integers: $max"
    ).join("\n")
  );
}

Yes, it looks like Perl.

$ raku/ch-1.raku
Example 1:
Input: @arr = (-3, 1, 2, -1, 3, -2, 4)
Output: 4

Count of positive integers: 4
Count of negative integers: 3
Maximum of count of positive and negative integers: 4

Example 2:
Input: @arr = (-1, -2, -3, 1)
Output: 3

Count of positive integers: 1
Count of negative integers: 3
Maximum of count of positive and negative integers: 3

Example 3:
Input: @arr = (1, 2)
Output: 2

Count of positive integers: 2
Count of negative integers: 0
Maximum of count of positive and negative integers: 2

View the entire Raku script for this task on GitHub.

Perl

In Perl, we can get max and sum from List::Util.

sub maxPosNeg(@ints) {
  my $pos = sum map { $_ > 0 ? 1 : 0 } @ints;
  my $neg = sum map { $_ < 0 ? 1 : 0 } @ints;
  my $max = max $pos, $neg;
  return (
    $max,
    join("\n",
      "Count of positive integers: $pos",
      "Count of negative integers: $neg",
      "Maximum of count of positive and " .
      "negative integers: $max"
    )
  );
}

View the entire Perl script for this task on GitHub.

Python

For Python, sum and max are built in and don’t need to be pulled in from a library.

def maxPosNeg(ints):
    pos = sum([1 for i in ints if i > 0])
    neg = sum([1 for i in ints if i < 0])
    maxCount = max(pos, neg)
    return (
      maxCount,
      "\n".join([
          f"Count of positive integers: {pos}",
          f"Count of negative integers: {neg}",
          f"Maximum of count of positive and " +
          f"negative integers: {maxCount}"
      ])
    )

View the entire Python script for this task on GitHub.

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Task 2: Count Equal Divisible

You are given an array of integers, @ints and an integer $k.

Write a script to return the number of pairs (i, j) where

a) 0 <= i < j < size of @ints
b) ints[i] == ints[j]
c) i x j is divisible by k

Example 1

Input: @ints = (3,1,2,2,2,1,3) and $k = 2
Output: 4

(0, 6) => ints[0] == ints[6] and 0 x 6 is divisible by 2
(2, 3) => ints[2] == ints[3] and 2 x 3 is divisible by 2
(2, 4) => ints[2] == ints[4] and 2 x 4 is divisible by 2
(3, 4) => ints[3] == ints[4] and 3 x 4 is divisible by 2

Example 2

Input: @ints = (1,2,3) and $k = 1
Output: 0

Approach

Ok, let’s look at these criteria:

0 <= i < j < size of @ints. For a 0-indexed array, it means that both i and j are indices of the array (the 0 <= and < size of @ints parts) and that i < j. Not a big deal.

ints[i] == ints[j] means the numbers at these indices are the same. So Example 2 fails this criteria because none of the numbers are the same.

i x j is divisible by k. Really, this is the big condition.

As with the last task, we’re counting.

Raku

Here I’m leaning into the Raku looking like Perl.

sub countEquDiv($k, @ints) {
  my @explain;
  my $cnt = 0;
  for 0 .. @ints.end - 1 -> $i {
    for $i + 1 .. @ints.end -> $j {
      # does ints[i] == ints[j]?
      next unless @ints[$i] == @ints[$j];
      # is i x j divisible by k?
      next unless ( ($i * $j) mod $k ) == 0;
      # count the pair and explain why
      $cnt++;
      @explain.push(
        "($i, $j) => ints[$i] == ints[$j] " ~
        "and $i x $j is divisible by $k"
      );
    }
  }
  return($cnt, @explain.join("\n"));
}

$ raku/ch-2.raku
Example 1:
Input: @arr = (3, 1, 2, 2, 2, 1, 3) and $k = 2
Output: 4

(0, 6) => ints[0] == ints[6] and 0 x 6 is divisible by 2
(2, 3) => ints[2] == ints[3] and 2 x 3 is divisible by 2
(2, 4) => ints[2] == ints[4] and 2 x 4 is divisible by 2
(3, 4) => ints[3] == ints[4] and 3 x 4 is divisible by 2

Example 2:
Input: @arr = (1, 2, 3) and $k = 1
Output: 0

View the entire Raku script for this task on GitHub.

Perl

sub countEquDiv($k, @ints) {
  my @explain;
  my $cnt = 0;
  foreach my $i ( 0 .. $#ints - 1 ) {
    foreach my $j ( $i + 1 .. $#ints ) {
      # does ints[i] == ints[j]?
      next unless $ints[$i] == $ints[$j];
      # is i x j divisible by k?
      next unless ( ($i * $j) % $k ) == 0;
      # count the pair and explain why
      $cnt++;
      push @explain,
        "($i, $j) => ints[$i] == ints[$j] " .
        "and $i x $j is divisible by $k";
    }
  }
  return($cnt, join("\n", @explain));
}

View the entire Perl script for this task on GitHub.

Python

Here I’m leaning into the Python looking like Perl. I mean, except for the lack of block delimiters and sigils, how can you NOT think this looks like perl?

def countEquDiv(k, ints):
    explain = []
    cnt = 0
    for i in range(len(ints) - 1):
        for j in range(i+1, len(ints)):
            # does ints[i] == ints[j]?
            if not ints[i] == ints[j]: break
            # is i x j divisible by k?
            if not ( (i * j) % k ) == 0: break
            # count the pair and explain why
            cnt += 1
            explain.append(
                f"({i}, {j}) => ints[{i}] == ints[{j}] " +
                f"and {i} x {j} is divisible by {k}"
            )
    return(cnt, "\n".join(explain))

View the entire Python script for this task on GitHub.

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Here’s all my solutions in GItHub: https://github.com/packy/perlweeklychallenge-club/tree/master/challenge-262/packy-anderson