PWC228 - Unique Sum


Here we are with TASK #1 from The Weekly Challenge #228. Enjoy!

The challenge

You are given an array of integers.

Write a script to find out the sum of unique elements in the given array.

Example 1

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

In the given array we have 2 unique elements (1, 3).

Example 2

Input: @int = (1, 1, 1, 1)
Output: 0

In the given array no unique element found.

Example 3

Input: @int = (2, 1, 3, 4)
Output: 10

In the given array every element is unique.

The questions

What are the limits of the inputs, in terms of values and how many of them could appear in the array?

The solution

The questions can be important for the implementation. We will do like this:

  • track the result, starting from 0;
  • when a number appears the first time, we add it to the result;
  • when it appears the second time, it is subtracted;
  • any other time it appears, it is ignored.

This means that the accumulated, intermediate result might grow a lot big even though the final result might be much smaller.

Anyway, let’s move on with Raku:

#!/usr/bin/env raku
use v6;
sub MAIN (*@args) { put unique-sum(@args) }

sub unique-sum (@ints) {
   my $retval = 0;
   my $seen =;
   for @ints -> $x {
      given $seen{$x}++ {
         $retval += $x when 0;
         $retval -= $x when 1;
   return $retval;

We’re using a BagHash here because it’s kind of the right data structure, but any hash would do as we can see in the Perl alternative:

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

say unique_sum(@ARGV);

sub unique_sum (@ints) {
   my $retval = 0;
   my %seen;
   for my $x (@ints) {
      my $previous = $seen{$x}++ // 0;
      $retval += $x unless $previous;
      $retval -= $x if $previous == 1;
   return $retval;

The alternative solution might be to first filter out all elements with a duplicate, then do the sum. This means allocating possibly a lot of space… so there we go, it’s either/or. Well, unless a different algorithm can be found, of course!

Stay safe and cheers!

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