Here is an interesting coding challenge: write a counter function that counts from 1 to max but only returns numbers whose digits don't repeat.

...

Also:

- Display the biggest jump (in the sequences above, it's 4: 98 -> 102)
- Display the total count of numbers
- Give these two values for max=10000

In the comments section of his blog, you can find many solutions, some are long, some are short but have lovely punctuation soup like

`!/(.).*\1/, 1..MAX;`

(that's from the Perl solution).Here is a Factor version that is shorter than anything in the comments there:

USING: kernel math.parser math.ranges math.vectors sequences sets ;

1 10000 [a,b] [ number>string all-unique? ] filter

[ ] [ length ] [ dup 0 prefix v- supremum ] tri

It leaves three values on the stack (hence

`tri`

): a sequence of numbers with non-repeating digits, the length of this sequence, and the largest jump between any two numbers in the sequence.Of course one could claim that this is cheating, since I'm using a lot of library words here. Fair enough, but those library words in the library for a reason: they get used a hell of a lot.

## 2 comments:

Although having more tokens I prefer this version because it doesn't create an uneccesary copy of the the 5274 element sequence returned by filter by appending the prefix 0.

1 10000 [a,b] [ number>string all-unique? ] filter

[ ] [ length ] [ [ 1 tail-slice ] keep v- supremum ] tri

Crest: here's another approach which avoids both intermediate sequences:

1 10000 [a,b] [ number>string all-unique? ] filter

[ ] [ length ] [ 2 <clumps> 0 [ first2 swap - max ] reduce ] tri

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