RogerBW's Blog

The Weekly Challenge 234: Triplets with Character 17 September 2023

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

Task 1: Common Characters

You are given an array of words made up of alphabetic characters only.

Write a script to return all alphabetic characters that show up in all words including duplicates.

This fell for me into three major steps. In Perl:

sub commoncharacters ($a) {

Step 1: make a list ac of hashes of character counts.

  my @ac;
  foreach my $w (@{$a}) {
    my %h;
    foreach my $c (split '', $w) {
    push @ac, \%h;

Step 2: make a single hash vc of minimum character counts (no key if the character is not present in all words, otherwise the lowest number of times it occurs).

  my %vc = %{$ac[0]};
  foreach my $aa (@ac) {
    foreach my $c (keys %vc) {
      if (exists $aa->{$c}) {
        $vc{$c} = min($vc{$c}, $aa->{$c});
      } else {
        delete $vc{$c};

Step 3: for each character in the (arbitrarily chosen) first input word, output it if it's present in vc and decrement its count there, deleting the key once we've run out of that character to simplify the testing next time it comes up (e.g. the second "a" in "java").

  my @out;
  foreach my $c (split '', $a->[0]) {
    if (exists $vc{$c}) {
      push @out, $c;
      if ($vc{$c} == 0) {
        delete $vc{$c};
  return \@out;

In Rust I could use counter::Counter which has an overloaded & to produce a minimum-count result for step 2, so the whole thing becomes more compact.

fn commoncharacters(a: Vec<&str>) -> Vec<char> {
    let ac =
        a.iter().map(|w| w.chars().collect::<Counter<_>>()).collect::<Vec<_>>();
    let mut vc = ac[0].clone();
    for tc in ac.iter().skip(1) {
        vc = vc & tc.clone();
    let mut out = Vec::new();
    for c in a[0].chars() {
        if vc.contains_key(&c) {

Task 2: Unequal Triplets

You are given an array of positive integers.

Write a script to find the number of triplets (i, j, k) that satisfies num[i] != num[j], num[j] != num[k] and num[k] != num[i].

I decided to have some fun with this one. The naïve way would be to generate all valid combinations of (i, j, k) and check for inequality. But there's a better way. In Python:

def unequaltriplets(a):

Make a hash of numbers and their counts.

  utc = defaultdict(lambda: 0)
  for n in a:
    utc[n] += 1

Make a list of distinct numbers, and bail out if there will be no triplets.

  kl = list(utc.keys())
  if len(kl) < 3:
    return 0

Start the count.

  out = 0

For each triplet of distinct numbers:

  for c in combinations(kl, 3):

Increment the count by the product of the number of times each number occurs, which is the total number of triplets that can be formed. So if I had an input of [1, 1, 2, 2, 3] my hash of counts would be {1 => 2, 2 => 2, 3 => 1} and I'd increment by 2×2×1 = 4.

    out += utc[c[0]] * utc[c[1]] * utc[c[2]]
  return out

Full code on github.

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