# ETOOBUSY đźš€ minimal blogging for the impatient

# PWC196 - Pattern 132

**TL;DR**

Here we are with TASK #1 from The Weekly Challenge #196. Enjoy!

# The challenge

You are given a list of integers,

`@list`

.Write a script to find out subsequence that respect

`Pattern 132`

. Return empty array if none found.Pattern 132 in a sequence (a[i], a[j], a[k]) such that

`i < j < k and a[i] < a[k] < a[j]`

.

Example 1`Input: @list = (3, 1, 4, 2) Output: (1, 4, 2) respect the Pattern 132.`

Example 2`Input: @list = (1, 2, 3, 4) Output: () since no susbsequence can be found.`

Example 3`Input: @list = (1, 3, 2, 4, 6, 5) Output: (1, 3, 2) if more than one subsequence found then return the first.`

Example 4`Input: @list = (1, 3, 4, 2) Output: (1, 3, 2)`

# The questions

I guess there are no questions to be asked, apart the old ones I donâ€™t ask any more like what to do if inputs are missing (at least three numbers are needed), or not compliant with the constraints.

# The solution

Weâ€™ll tackle this in the plain, old brute-forceish way, i.e. a triply
nested loop. Thereâ€™s a little optimization when we look for â€śhighâ€ť
middle values that should be big enough to be both greater than the
â€ślowâ€ť counterpart, as well as leave some intermediate space for the
â€śmidâ€ť one. Nothing that makes life *less complex*.

```
#!/usr/bin/env raku
use v6;
sub MAIN (*@args is copy) {
@args = 3, 1, 4, 2 unless @args;
say first-pattern132(@args);
}
sub first-pattern132 (@list) {
for 0 .. (@list - 3) -> $low {
for $low + 1 .. (@list - 2) -> $high {
next if @list[$high] <= @list[$low] - 1;
for $high + 1 ..^ @list -> $mid {
return @list[$low, $high, $mid]
if @list[$low] < @list[$mid] < @list[$high];
}
}
}
return ();
}
```

The translation in Perl is straightforward, because Iâ€™m not using any Raku magic in the solution above.

```
#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';
my @result = first_pattern132(@ARGV ? @ARGV : (3, 1, 4, 2));
say "(@result)";
sub first_pattern132 (@list) {
for my $low (0 .. (@list - 3)) {
for my $high ($low + 1 .. (@list - 2)) {
next if $list[$high] <= $list[$low] - 1;
for my $mid ($high + 1 .. (@list - 1)) {
return @list[$low, $high, $mid]
if $list[$low] < $list[$mid] && $list[$mid] < $list[$high];
}
}
}
return;
}
```

This saidâ€¦ all is said, cheers!

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