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| -rw-r--r-- | challenge-306/jeanluc2020/blog-1.txt | 1 | ||||
| -rw-r--r-- | challenge-306/jeanluc2020/blog-2.txt | 1 | ||||
| -rwxr-xr-x | challenge-306/jeanluc2020/perl/ch-1.pl | 65 | ||||
| -rwxr-xr-x | challenge-306/jeanluc2020/perl/ch-2.pl | 116 |
4 files changed, 183 insertions, 0 deletions
diff --git a/challenge-306/jeanluc2020/blog-1.txt b/challenge-306/jeanluc2020/blog-1.txt new file mode 100644 index 0000000000..07a76c8dcd --- /dev/null +++ b/challenge-306/jeanluc2020/blog-1.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-306-1.html diff --git a/challenge-306/jeanluc2020/blog-2.txt b/challenge-306/jeanluc2020/blog-2.txt new file mode 100644 index 0000000000..5e234e59e6 --- /dev/null +++ b/challenge-306/jeanluc2020/blog-2.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-306-2.html diff --git a/challenge-306/jeanluc2020/perl/ch-1.pl b/challenge-306/jeanluc2020/perl/ch-1.pl new file mode 100755 index 0000000000..0676779cbf --- /dev/null +++ b/challenge-306/jeanluc2020/perl/ch-1.pl @@ -0,0 +1,65 @@ +#!/usr/bin/env perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-306/#TASK1 +# +# Task 1: Odd Sum +# =============== +# +# You are given an array of positive integers, @ints. +# +# Write a script to return the sum of all possible odd-length subarrays of the +# given array. A subarray is a contiguous subsequence of the array. +# +## Example 1 +## +## Input: @ints = (2, 5, 3, 6, 4) +## Output: 77 +## +## Odd length sub-arrays: +## (2) => 2 +## (5) => 5 +## (3) => 3 +## (6) => 6 +## (4) => 4 +## (2, 5, 3) => 10 +## (5, 3, 6) => 14 +## (3, 6, 4) => 13 +## (2, 5, 3, 6, 4) => 20 +## +## Sum => 2 + 5 + 3 + 6 + 4 + 10 + 14 + 13 + 20 => 77 +# +## Example 2 +## +## Input: @ints = (1, 3) +## Output: 4 +# +############################################################ +## +## discussion +## +############################################################ +# +# We walk the array from the first element to the last and +# add the sum of all subarrays starting at the current element +# when the number of elements in that subarray is odd. + +use strict; +use warnings; +use List::Util qw(sum); + +odd_sum(2, 5, 3, 6, 4); +odd_sum(1, 3); + +sub odd_sum { + my @ints = @_; + print "Input: (" . join(",", @ints) . ")\n"; + my $sum = 0; + foreach my $i (0..$#ints) { + foreach my $j ($i..$#ints) { + my @tmp = @ints[$i..$j]; + if(scalar(@tmp) % 2) { + $sum += sum(@tmp); + } + } + } + print "Output: $sum\n"; +} diff --git a/challenge-306/jeanluc2020/perl/ch-2.pl b/challenge-306/jeanluc2020/perl/ch-2.pl new file mode 100755 index 0000000000..49b768f2ed --- /dev/null +++ b/challenge-306/jeanluc2020/perl/ch-2.pl @@ -0,0 +1,116 @@ +#!/usr/bin/env perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-306/#TASK2 +# +# Task 2: Last Element +# ==================== +# +# You are given a array of integers, @ints. +# +# Write a script to play a game where you pick two biggest integers in the +# given array, say x and y. Then do the following: +# +# a) if x == y then remove both from the given array +# b) if x != y then remove x and replace y with (y - x) +# +# At the end of the game, there is at most one element left. +# +# Return the last element if found otherwise return 0. +# +## Example 1 +## +## Input: @ints = (3, 8, 5, 2, 9, 2) +## Output: 1 +## +## Step 1: pick 8 and 9 => (3, 5, 2, 1, 2) +## Step 2: pick 3 and 5 => (2, 2, 1, 2) +## Step 3: pick 2 and 1 => (1, 2, 2) +## Step 4: pick 2 and 1 => (1, 2) +## Step 5: pick 1 and 2 => (1) +# +## Example 2 +## +## Input: @ints = (3, 2, 5) +## Output: 0 +## +## Step 1: pick 3 and 5 => (2, 2) +## Step 2: pick 2 and 2 => () +# +############################################################ +## +## discussion +## +############################################################ +# +# There is a bit of ambiguity in the description that is clarified +# by the examples: if there are multiple elements of different value, +# but also multiple elements that are as big as the maximum element, +# then pick one of the biggest values and one of the second biggest, +# NOT two of the biggest. So from (2, 2, 1, 2) pick 2 and 1, not +# 2 and 2. +# With that clarified, we can now implement the whole thing: Search +# the two biggest values in the array, then handle them as layed out +# in the rules. +# + +use v5.36; + +last_element(3, 8, 5, 2, 9, 2); +last_element(3, 2, 5); + +sub last_element(@ints) { + say "Input: (" . join(",", @ints) . ")"; + while(@ints) { + if(1 == scalar(@ints)) { + say "Output: " . $ints[0]; + return; + } + my ($max1, $max2) = @ints[0..1]; + if($max1 < $max2) { + ($max1, $max2) = ($max2, $max1); + } + foreach my $i (2..$#ints) { + if($ints[$i] > $max1) { + ($max1, $max2) = ($ints[$i], $max1); + } elsif ($max1 == $max2) { + $max2 = $ints[$i]; + } else { + if($ints[$i] > $max2 && $ints[$i] < $max1) { + $max2 = $ints[$i]; + } + } + } + if($max1 == $max2) { + # remove two occurences of $max1 + my @tmp = (); + my $removed = 0; + foreach my $elem (@ints) { + if($elem == $max1 && $removed < 2) { + $removed++; + next; + } + push @tmp, $elem; + } + @ints = @tmp; + } else { + # remove max2, replace max1 with max1-max2 + my @tmp = (); + my $new = $max1 - $max2; + my $removed = 0; + my $replaced = 0; + foreach my $elem (@ints) { + if($elem == $max1 && ! $removed) { + $removed = 1; + next; + } + if($elem == $max2 && ! $replaced) { + push @tmp, $new; + $replaced = 1; + next; + } + push @tmp, $elem; + } + @ints = @tmp; + } + } + say "Output: 0"; +} |