diff options
| -rw-r--r-- | challenge-052/mark-anderson/raku/ch-1.p6 | 57 |
1 files changed, 55 insertions, 2 deletions
diff --git a/challenge-052/mark-anderson/raku/ch-1.p6 b/challenge-052/mark-anderson/raku/ch-1.p6 index 1bbc8648fe..1ffb7ccd0e 100644 --- a/challenge-052/mark-anderson/raku/ch-1.p6 +++ b/challenge-052/mark-anderson/raku/ch-1.p6 @@ -3,11 +3,64 @@ # I learned about polymod from Kevin Colyer's ch-2.p6 solution last week. # I like that method for splitting up the number into its digits. -sub MAIN($num1, $num2 where $num1 >= 100 < $num2 <= 999) { - for ($num1..$num2) -> $num { +sub MAIN($arg1, $arg2 where $arg1 >= 100 < $arg2 <= 999) { + for ($arg1..$arg2) -> $num { my @digits = (+$num).polymod(10,10); if (([-] @digits[0,1]).abs == ([-] @digits[1,2]).abs == 1) { say $num; } } } + +# There is a breadth first search solution that I found at +# https://www.geeksforgeeks.org/stepping-numbers/ +# I translated the C++ code into raku below. This is an +# alternate solution to my usual brute force approach above. + +#`[ +my @results; + +sub MAIN($arg1, $arg2 where $arg1 >= 100 < $arg2 <= 999) { + for (0 .. 9) -> $num { + bfs($arg1, $arg2, $num); + } + + @results.sort.join("\n").say; +} + +sub bfs(Int $n, Int $m, Int $num) { + my @nums; + + @nums.push($num); + + while (@nums) { + my $stepNum = @nums.shift; + + if ($m >= $stepNum >= $n) { + @results.push($stepNum); + } + + if ($num == 0 || $stepNum > $m) { + next; + } + + my $lastDigit = $stepNum % 10; + + my $stepNumA = $stepNum * 10 + ($lastDigit - 1); + my $stepNumB = $stepNum * 10 + ($lastDigit + 1); + + if ($lastDigit == 0) { + @nums.push($stepNumB); + } + + elsif ($lastDigit == 9) { + @nums.push($stepNumA); + } + + else { + @nums.push($stepNumA); + @nums.push($stepNumB); + } + } +} +] |