RogerBW's Blog

The Weekly Challenge 250: Smallest Viable Value 07 January 2024

I’ve been doing the Weekly Challenges. The latest involved searching sequences and parsing strings. (Note that this ends today.)

But first: many thanks to Mohammad Anwar for keeping this going (and writing most of the challenges). I've been doing this since number 16 and I'm still having a good time (even if I'm spreading it over ten different languages now).

Task 1: Smallest Index

You are given an array of integers, @ints.

Write a script to find the smallest index i such that i mod 10 == $ints[i] otherwise return -1.

Straightforward if an early return is possible, wihch in most places it is.

(I seem to have solved a slightly different problem, as I'm testing for $ints[i] mod 10 == i. But I'm not sure it works as specified anyway. Let my carelessness stand as a warning against yours.)


fun smallestindex(a: List<Int>): Int {
    a.forEachIndexed {i, n ->
                          if (n % 10 == i) {
                              return i
    return -1

Task 2: Alphanumeric String Value

You are given an array of alphanumeric strings.

Write a script to return the maximum value of alphanumeric string in the given array.

The value of alphanumeric string can be defined as

a) The numeric representation of the string in base 10 if it is made up of digits only.

b) otherwise the length of the string

In Perl, this is clearly a regexp check.

sub alphanumericstringvalue($a) {
  my @l;
  foreach my $n (@{$a}) {
    if ($n =~ /^[0-9]+$/) {
      push @l, 0 + $n;
    } else {
      push @l, length($n);
  return max(@l);

In most other languages, this is a matter of trying to conversion to integer and catching any error, whether by flagged return value (JavaScript):

function alphanumericstringvalue(a) {
    let l = [];
    for (let n of a) {
        let p = 1 * n;
        if (Number.isNaN(p)) {
            p = n.length;
    return Math.max(...l);

or with an actual exception handler (Python):

def alphanumericstringvalue(a):
  l = []
  for n in a:
    p = 0
      p = int(n)
      p = len(n)
  return max(l)

Full code on github.

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