TL;DR

On with TASK #2 from The Weekly Challenge #225. Enjoy!

The challenge

You are given an array of integers, @ints.

Write a script to return left right sum diff array as shown below:

@ints = (a, b, c, d, e)

@left  = (0, a, (a+b), (a+b+c))
@right = ((c+d+e), (d+e), e, 0)
@left_right_sum_diff = ( | 0 - (c+d+e) |,
                         | a - (d+e)   |,
                         | (a+b) - e   |,
                         | (a+b+c) - 0 | )

Example 1:

Input: @ints = (10, 4, 8, 3)
Output: (15, 1, 11, 22)

@left  = (0, 10, 14, 22)
@right = (15, 11, 3, 0)

@left_right_sum_diff = ( |0-15|, |10-11|, |14-3|, |22-0|)
                     = (15, 1, 11, 22)

Example 2:

Input: @ints = (1)
Output: (0)

@left  = (0)
@right = (0)

@left_right_sum_diff = ( |0-0| ) = (0)

Example 3:

Input: @ints = (1, 2, 3, 4, 5)
Output: (14, 11, 6, 1, 10)

@left  = (0, 1, 3, 6, 10)
@right = (14, 12, 9, 5, 0)

@left_right_sum_diff = ( |0-14|, |1-12|, |3-9|, |6-5|, |10-0|)
                     = (14, 11, 6, 1, 10)

The questions

I mean… the initial inspiration is wrong, I assume, because the array has five elements but the output only contains four, which seems to be contradicted by every single example provided. Hence, I guess that the actual specification’s example should be:

@ints = (a, b, c, d, e)

@left  = (0, a, (a+b), (a+b+c), (a+b+c+d))
@right = ((b+c+d+e), (c+d+e), (d+e), e, 0)
@left_right_sum_diff = ( | 0 - (b+c+d+e) |,
                         | a   - (c+d+e) |,
                         | (a+b) - (d+e) |,
                         | (a+b+c)   - e |,
                         | (a+b+c+d) - 0 |)

Anyway!

The solution

Well, first of all congratulations to manwar for his well-deserved achievement. I couldn’t think of someone more deserving of the White Camel Award, especially in the year that it transitions from three people down to one.

I only have two regrets regarding this: the video does not show him properly, and I would have liked to congratulate with him in person.

OK, let’s get back to (his) business. We will pass through the array twice: on the first we will accumulate values, building what amounts to the @left array in the examples; on the second, we will subtract values in-place as we go.

Perl first:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';

{ local $" = ', '; say "(@{[ left_right_sum_diff(@ARGV) ]})" }

sub left_right_sum_diff (@input) {
   my $sum = 0;
   my @retval;
   for my $i (0 .. $#input) {
      push @retval, $sum;
      $sum += $input[$i];
   }
   for my $i (0 .. $#input) {
      $sum -= $input[$i];
      my $diff = $retval[$i] - $sum;
      $retval[$i] = $diff > 0 ? $diff : -$diff;
   }
   return @retval;
}

Raku:

#!/usr/bin/env raku
use v6;
sub MAIN (*@inputs) { say left-right-sum-diff(@inputs) }

sub left-right-sum-diff (@inputs) {
   my $sum = 0;
   my @retval;
   for ^@inputs -> $i {
      @retval.push: $sum;
      $sum += @inputs[$i];
   }
   for ^@inputs -> $i {
      $sum -= @inputs[$i];
      @retval[$i] = abs(@retval[$i] - $sum);
   }
   return @retval;
}

And… that’s all folks!