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| -rw-r--r-- | challenge-194/james-smith/README.md | 14 |
1 files changed, 10 insertions, 4 deletions
diff --git a/challenge-194/james-smith/README.md b/challenge-194/james-smith/README.md index 5480e2db42..579581f98f 100644 --- a/challenge-194/james-smith/README.md +++ b/challenge-194/james-smith/README.md @@ -58,15 +58,21 @@ sub digit_2400 { ## Solution -Again we use and IIFE. This time taking the value of the sorted values from `%sc`. For the method to be true. The sorted values must be: +Again we use and IIFE. This time taking the value of the sorted values from `%f` (being the frequencies of each letter). +For the method to be true. The sorted values must be: * `n`, `n`, `n`, ...., `n`, `n+1` -So we just need to check the 1st with the (n-1)st and the (n-1)st with the last! +In fact if we reverse the sort (switch `$a` and `$b` around in the comparison) we have: + + * `n+1`, `n`, `n`, ...., `n`, `n` + +So check to see if the first is one more than the second and the second is the same +as the last. ```perl sub check { - my %sc; $sc{$_} ++ for split //,$_[0]; - sub { @_>2 && $_[0]==$_[1]+1 && $_[-1]==$_[1] ? 1 : 0 }->(sort {$b<=>$a} values %sc); + my %f; $f{$_} ++ for split //, $_[0]; + sub { @_>2 && $_[0]==$_[1]+1 && $_[-1]==$_[1] }->(sort {$b<=>$a} values %f) || 0; } ``` |