TL;DR

On with TASK #2 from The Weekly Challenge #208. Enjoy!

# The challenge

You are given an array of integers in sequence with one missing and one duplicate.

Write a script to find the duplicate and missing integer in the given array. Return -1 if none found.

For the sake of this task, let us assume the array contains no more than one duplicate and missing.

Example 1:

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

Duplicate is 2 and Missing is 3.


Example 2:

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

No duplicate and missing found.


Example 3:

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

Duplicate is 3 and Missing is 4.


# The questions

This issue sparked so many questions that there must be something wrong somewhere. I mean, either I’ve become increasingly soft lately, sparing our fine host manwar of my usual useless nitpicking, or he decided to throw me a bone just to have a good laugh.

Anyway.

All examples seem to start at 1. Is this a general rule? It’s not in the rules, so I guess it’s a coincidence and will not assume this.

Is the “sequence” in the input array supposed to be an arithmetic progression with common difference equal to 1? I mean, any sequence is a sequence 🙄

The last example seems to imply that not finding the missing element within the array is still OK, because it must be the one immediately following. Why not the one immediately before (like 0 in the example)?

How are we supposed to return/print the two elements, if present? Should it always be duplicate first, then missing? Or in the order of their detection (which also happens to be their order of discovery, assuming that the missing missing is the one immediately after)?

Is there any assumption that can be done about the position of the duplicates and missing elements? I mean, are their positions totally random and not correlated to one another, or anything else?

Is the sequence mostly short, or should we cope with very long sequences?

# The solution

The last two questions come from my strong tendency to over-engineer. I mean…

• if the inputs are very long…
• … and the distribution of the duplicates and missing are random and not correlated…

it might make sense to think about optimizing the solution with some adaptation of a binary search and linear search below a certain point.

Why “not correlated”? If the missing always occurs very close to the duplicate, it’s very hard to spot the division points by binary search, so it would just make sense to do a linear search.

Anyway, let’s assume it remains at the toy level, so we will only focus on the linear search, right?

Let’s start with Perl first. We can assume that the input is OK, but given all other questions… can we trust these assumptions?!? Let the user (partially) decide!

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

our $SHORT_CIRCUIT =$ENV{SHORT_CIRCUIT} // 1;
if (my @dam = duplicate_and_missing(@ARGV)) {
local $" = ','; say "(@dam)"; } else { say -1; } sub duplicate_and_missing (@list) { my ($duplicate, $missing, @retval); for my$i (1 .. $#list) { if ($list[$i] ==$list[$i - 1]) { die "too many duplicates ($duplicate, $list[$i])\n"
if defined $duplicate; push @retval,$duplicate = $list[$i];
}
elsif ($list[$i] == $list[$i - 1] + 2) {
my $miss =$list[$i] - 1; die "too many missing ($missing, $miss)\n" if defined$missing;
push @retval, $missing =$miss;
}
elsif ($list[$i] != $list[$i - 1] + 1) {
die "unexpected gap\n";
}
else {} # just a simple increment
return @retval if @retval == 2 && our $SHORT_CIRCUIT; } return unless defined($duplicate);
push @retval, $list[-1] + 1 unless @retval == 2; return @retval; }  The Raku version can appear to be… disappointing. I nknow there must be more idiomatic ways of putting it, but using them at every cost might be bad for my readability, so let’s not pull the rope too much! #!/usr/bin/env raku use v6; sub MAIN (*@args) { my @dam = duplicate-and-missing(@args, %*ENV<SHORT_CIRCUIT>); put @dam ?? "({@dam.join(',')})" !! -1; } sub duplicate-and-missing (@list,$short-circuit is copy = Nil) {
$short-circuit //= True; my ($duplicate, $missing, @retval); for 1 ..^ @list ->$i {
if @list[$i] == @list[$i - 1] {
die "too many duplicates ($duplicate, {@list[$i]})\n"
if defined $duplicate; @retval.push:$duplicate = @list[$i]; } elsif (@list[$i] == @list[$i - 1] + 2) { my$miss = @list[$i] - 1; die "too many missing ($missing, $miss)\n" if defined$missing;
@retval.push: $missing =$miss;
}
elsif (@list[$i] != @list[$i - 1] + 1) {
die "unexpected gap\n";
}
else {} # just a simple increment
return @retval if @retval == 2 && $short-circuit; } return [] unless defined($duplicate);
@retval.push(@list[*-1] + 1) unless @retval == 2;
return @retval;
}


I guess it’s everything for this post, stay safe folks!