RogerBW's Blog: The Weekly Challenge 378: Second Sum

I’ve been doing the Weekly Challenges. The latest involved . (Note that this ends today.)

Task 1: Second Largest Digit

You are given an alphanumeric string.

Write a script to find the second largest distinct digit in the given string. Return -1 if none found.

So: find the digits, make a list of distinct ones, return the second largest. In Perl:

sub secondlargestdigit($a) {

Split the string, find the digits, and put them into a pseudo-set.

  my %p = map {$_ => 1} grep /[0-9]/, split '', $a;

Sort the entries of that pseudo-set in descending order.

  my @k = reverse sort keys %p;

If we have enough of them, return the second, otherwise -1.

  if (scalar @k > 1) {
    $k[1];
  } else {
    -1;
  }
}

PostScript is pleasingly straightforward (though it relies on some of my library functions).

/secondlargestdigit {

Get a list.

s2a

Filter it for digits.

    {
        dup 48 ge exch 57 le and
    } filter

Take just the unique digits.

    toset keys

Sort in reverse order.

    quicksort reverse

If it's long enough, return the value of the second digit, otherwise -1.

    dup length 1 gt {
        1 get 48 sub
    } {
        pop -1
    } ifelse
} bind def

Task 2: Sum of Words

You are given three strings consisting of lower case English letters "a" to "j" only. The letter value of a = 0, b = 1, c = 3, etc.

Write a script to find if sum of first two strings return the third string.

The examples of leading "a" (i.e. leading zero) make it clear that there are multiple allowable representations of any given number. Therefore I write only one direction of conversion, from this representation to an integer, and then compare integer values.

In Lua:

function l2n(a)

Establish a reference value for "a".

   local a0 = string.byte("a")

Initialise the accumulator.

   local t = 0

For each "digit",

   for _, c in ipairs(split(a)) do

Multiply the previous accumulator value by 10, and add the new digit's value.

      t = t * 10 + string.byte(c) - a0
   end
   return t
end

Then the target function becomes trivial:

function sumofwords(a, b, c)
   return l2n(a) + l2n(b) == l2n(c)
end

Again I find the PostScript pleasing (and I'm always glad not to have to use local variables).

/l2n {

Put a zero accumulator on the stack below the input string.

    0 exch

For each character of that string,

    s2a {

Swap with the accumulator, multiply it by 10, then add character and subtract "a" (PostScript only works in ASCII as far as I know, certainly on any system I have access to),

        exch
        10 mul
        add
        97 sub
    } forall
} bind def

Then it's just a matter of converting all the inputs and checking for equality.

/sumofwords {
    3 1 roll
    l2n exch
    l2n exch
    add
    exch
    l2n eq
} bind def

Full code on codeberg.

RogerBW's Blog

Add A Comment