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| -rw-r--r-- | challenge-058/luca-ferrari/raku/ch-2.p6 | 49 |
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diff --git a/challenge-058/luca-ferrari/raku/ch-2.p6 b/challenge-058/luca-ferrari/raku/ch-2.p6 new file mode 100644 index 0000000000..656be20648 --- /dev/null +++ b/challenge-058/luca-ferrari/raku/ch-2.p6 @@ -0,0 +1,49 @@ +#!env raku +# +# Perl Weekly Challenge 58 +# +# Task 2 +# +# Write a script to arrange people in a lineup according to how many taller people are in front of each person in line. You are given two arrays. @H is a list of unique heights, in any order. @T is a list of how many taller people are to be put in front of the corresponding person in @H. The output is the final ordering of people’s heights, or an error if there is no solution. + +# Here is a small example: + +# @H = (2, 6, 4, 5, 1, 3) # Heights +# @T = (1, 0, 2, 0, 1, 2) # Number of taller people in front + +# The ordering of both arrays lines up, so H[i] and T[i] refer to the same person. For example, +# there are 2 taller people in front of the person with height 4, and there is 1 person in front of the person with height 1. +# +# As per the last diagram, your script would then output the ordering (5, 1, 2, 6, 3, 4) in this case. (The leftmost element is the “front” of the array.) + +sub MAIN(){ + my @H = 2, 6, 4, 5, 1, 3; + my @T = 1, 0, 2, 0, 1, 2; + + + # build an hash to map heights to tallers + my %HT; + for 0 ..^ @H.elems { + %HT{ @H[ $_ ] } = @T[ $_ ]; + } + + + # evaluate all possible solutions + for %HT.keys.permutations -> @solution { + # the leftmost element must have a taller set to zero! + next if %HT{ @solution[ 0 ] } != 0; + + # the solution is good if the number of tallers for + # every element is equal to the values of tallers + my $ok = True; + + for 1 ..^ @solution.elems { + my $height = @solution[ $_ ]; + my $tallers = %HT{ $height }; + $ok = False if @solution[ 0 .. $_-1].grep( * > $height ) != $tallers; + last if ! $ok; + } + + say @solution if $ok; + } +} |