RogerBW's Blog

The Weekly Challenge 194: Digital Equaliser 11 December 2022

I’ve been doing the Weekly Challenges. The latest involved expanding a search space and running a character frequency test. (Note that this closes today.)

Task 1: Digital Clock

You are given time in the format hh:mm with one missing digit.

Write a script to find the highest digit between 0-9 that makes it valid time.

Wild digits 3 and 4 always give the same result. Wild digits 1 and 2 vary in their result depending on the value of the other digit in the pair. And that's about it. Kotlin:

fun digitalclock(hhmm: String): Int {
    val i = hhmm.indexOf("?")
    if (i == 0) {
        if (hhmm[1] <= '3') {
            return 2;
        return 1;
    } else if (i == 1) {
        if (hhmm[0] < '2') {
            return 9;
        return 3;
    } else if (i == 3) {
        return 5;
    } else if (i == 4) {
        return 9;
    return 0;

Task 2: Frequency Equalizer

You are given a string made of alphabetic characters only, a-z.

Write a script to determine whether removing only one character can make the frequency of the remaining characters the same.

To avoid chugging through with trial and error:

  1. Take a list of the frequency of occurrence of each character.

  2. Sort it numerically.

  3. Then the test is true if and only if the first value is the same as the second-to-last, and one less than the last.


sub frequencyequalizer($s) {
  my %f;
  for $s.comb -> $c {
  my @v = %f.values.sort;
  if (@v[0] == @v[*-2] &&
      @v[0] + 1 == @v[*-1]) {
    return True;
  return False;

Full code on github.

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