# RogerBW's Blog

 The Weekly Challenge 230: The Count of Separation 20 August 2023 I’ve been doing the Weekly Challenges. The latest involved number lists and word prefixes. (Note that this ends today.) Task 1: Separate Digits You are given an array of positive integers. Write a script to separate the given array into single digits. This is a pretty easy one to convert into an algorithm: break down each number into digits, and append digits to the output list. Perl: ``````sub separatedigits(\$a) { my @out; `````` Take each input number in turn. `````` foreach my \$n (@{\$a}) { `````` Make a working copy. `````` my \$m = \$n; `````` Set up this number's output list. `````` my @v; `````` Specifying "positive" rather than "non-negative" integers makes my life easier here; I can ignore the special case where the initial value of `\$m` is zero. `````` while (\$m > 0) { `````` Push the mod10 value onto the list and divide the working value. `````` push @v, \$m % 10; \$m = int(\$m / 10); } `````` Now I have a list of the digits in reverse order (least significant first) – so reverse that list and append it to the output. `````` push @out, reverse @v; } return \@out; } `````` Other languages are broadly similar. Ruby has its `divmod` method which is probably slightly more efficient than separate division and modulus operations. Lua doesn't have a list reverser so I end up prepending values to `v`. Every language has a different way of saying "extend this list with the elements of this other list". Task 2: Count Words You are given an array of words made up of alphabetic characters and a prefix. Write a script to return the count of words that starts with the given prefix. This is the sort of problem I like to solve in a left-to-right functional style: take the list, filter by prefix, take the length. I couldn't get Raku to do this with a `grep`, I think because of the way its implementation of `index` can produce a NaN-like `Nil` value ("the substring is not in the string at all") which has to be checked separately from a non-zero numeric value ("the substring is in the string, just not at the front"). So I just laid it out in full: ``````sub prefixwords(@s, \$p) { my \$r = 0; for @s -> \$w { with \$w.index(\$p) -> \$i { if (\$i == 0) { \$r++; } } } return \$r; } `````` In several other languages this could be a one-liner, as in Kotlin: ``````fun prefixwords(s: List, p: String): Int { return s.filter {it.startsWith(p)}.size } `````` Ruby lets me be even briefer by having a filter-and-count keyword: ``````def prefixwords(s, p) return s.count {|i| i.index(p) == 0} end `````` I rather like the efficiency of PostScript: ``````/prefixwords { 1 dict begin /p exch def [ exch { `````` The `anchorsearch` keyword returns either `(string-after-match) (match) true` or `(string) false` – so by discarding the top two elements on the stack I end up with either the after-match or nothing at all. And since I'm only going to be counting the number of successful matches, I can casually leave that lying on the stack as the countable element. `````` p anchorsearch pop pop } forall ] length end } bind def `````` (I could have used a `mark` and `counttomark`/`cleartomark` – which would probably be slightly more efficient – but I'd need to store the result somewhere while clearing down the stack, and I like this approach of building up arrays.) Full code on github. Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one. 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