RogerBW's Blog

The Weekly Challenge 189: Degree of Character 06 November 2022

I’ve been doing the Weekly Challenges. The latest involved filtering character lists and finding subarrays. (Note that this is open until 6 November 2022.)

Task 1: Greater Character

You are given an array of characters (a..z) and a target character.

Write a script to find out the smallest character in the given array lexicographically greater than the target character.

We establish from the examples that if no character in the array is greater than the target, we return the target. Therefore:

sub greatercharacter($a, $k) {
  my @aa = sort grep {$_ gt $k} @{$a};
  if (@aa) {
    return $aa[0];
  } else {
    return $k;

In PostScript I use the convention that individual characters of strings are represented as integers and use those internally, converting back to a string at the end.

/greatercharacter {
    1 dict begin
    0 get /k exch def
    { 0 get } map { k gt } filter
    [ exch
      dup length 0 gt {
          quicksort 0 get
      } {
      } ifelse
    ] a2s
} bind def

(map, filter, quicksort and a2s are from my PostScript libraries.)

Task 2: Array Degree

You are given an array of 2 or more non-negative integers.

Write a script to find out the smallest slice, i.e. contiguous subarray of the original array, having the degree of the given array.

The degree of an array is the maximum frequence of an element in the array.

This is the sort of fiddly data structure manipulation that I mostly learned in Perl, and it's very good for me to work it out in other languages too. Overall:

  • I build a frequency map f of the values in the array. The highest value in that is the degree. I only care about numbers which occur that many times.

  • I build an inverse location table, telling me where each number is to be found. In example 5's [2, 1, 2, 1, 1], this would be {1 => [1, 3, 4], 2 => [0, 2]}. (Since I only care about first and last values, for some of the languages I make this explicitly a two-element list.)

  • I then look through all the numbers that occur (degree) times, and generate a length from inverse: position of the last occurrence, minus position of the first occurrence, plus one for the fencepost. If that length is smaller than the smallest length I've seen, store it and the relevant indices.

  • Finally return the array slice defined by those indices.

In Kotlin:

fun arraydegree(a: List<Int>): List<Int> {

Build the frequency map and calculate the degree:

    var f = mutableMapOf<Int, Int>()
    for (x in a) {
        if (f.containsKey(x)) {
            f[x] = f[x]!! + 1
        } else {
            f[x] = 1
    val degree = f.values.maxOrNull()!!

Build the inverse location table:

    var inverse = mutableMapOf<Int, MutableList<Int>>()
    for ((i, x) in a.withIndex()) {
        if (inverse.containsKey(x)) {
        } else {
            inverse[x] = mutableListOf(i)

Iterate through each number that exists in the correct degree, working out the length of the subarray including its first and last positions, and store those positions if it'll be shorter than any previously calculated length.

    var minlen = 1 + a.size
    var se = listOf(0, 1)
    for (n in f.keys) {
        if (f[n]!! == degree) {
            val ll = 1 + inverse[n]!![inverse[n]!!.size-1] - inverse[n]!![0]
            if (ll < minlen) {
                minlen = ll
                se = listOf(inverse[n]!![0], inverse[n]!![inverse[n]!!.size-1])

Finally return the array slice.

    return a.slice(se[0][1])

Full code on github.

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