RogerBW's Blog

The Weekly Challenge 228: The Empty Sum 06 August 2023

I’ve been doing the Weekly Challenges. The latest involved inspecting integer elements and making iterative things less so. (Note that this ends today.)

Task 1: Unique Sum

You are given an array of integers.

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

Fairly straightforward: convert the list into a hash with values equal to the count of elements. Raku has a Bag structure that's good for this, and Rust has the Counter crate that I've been using a great deal lately. Ruby, Kotlin and Python have hashes with defaults. Perl has autovivification, which is almost as good but rather less flexible. (Which just leaves Javascript, Lua and PostScript where I have to write it out the long way.)

Then take all the entries where count = 1, and sum them. Perl:

sub uniquesum($a) {
  my %c;
  map {$c{$_}++} @{$a};
  return sum0(grep {$c{$_} == 1} keys %c);

I like the explicitness of thw two stages in the Rust version.

fn uniquesum(a: Vec<u32>) -> u32 {

Break down the list into a Counter.

    let c = a.into_iter().collect::<Counter<u32>>();

Filter the object for count = 1, then output the values.

    c.iter().filter(|(_k, v)| **v == 1usize).map(|(k, _v)| k).sum()

Task 2: Empty Array

You are given an array of integers in which all elements are unique.

Write a script to perform the following operations until the array is empty and return the total count of operations.

If the first element is the smallest then remove it otherwise move it to the end.


I mean, I could do that. With a ring buffer or something.

Or I could say, well, we're basically counting the number of passes it takes to get the lowest value in the list to the front of the list, aren't we? So let's just work that out based on where it is in the list. And iterate for each number.

(There may be inputs for which this approach doesn't work. But it was fun to write.)

In Lua (where for once the 1-based list indices are actually useful):

function emptyarray(a0)
   local t = 0
   local a = a0
   while #a > 0 do

Find the minimum value.

      local mn = math.min(table.unpack(a))

Locate the minimum in the list.

      for i, v in ipairs(a) do
         if v == mn then

Remove it and count the number of "operations" needed to get here.

            table.remove(a, i)
            t = t + i
   return t

Full code on github.

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