RogerBW's Blog

Perl Weekly Challenge 74: more songs about lists 23 August 2020

I’ve been doing the Perl Weekly Challenges. The latest involved finding minima in subsets of lists.

1. Majority Element

You are given an array of integers of size $N.

Write a script to find the majority element. If none found then print -1.

Majority element in the list is the one that appears more than floor(sizeoflist/2).

Well, obviously I'll use max from List::Util.

sub majority {
  my @list=@_;
  my %s;

So first I break down the list into hash buckets: the bucket for N tells me how many times N appears.

  map {$s{$_}++} @list;

Now look at the highest value in a bucket. If it's higher than half the list size (as required in the problem statement), we have a majority element.

  my $m=max(values %s);
  if ($m > int((scalar @list)/2)) {

So then we invert the hash: key → value becomes value → key. (This will lose data in the case of duplicate values, but we don't care, because we're in the "is a majority element" case, so there can be only one hash bucket with a value of $m and that's the only one we'll be looking at.)

(Obviously this could be done iteratively – go through the hash keys and see which one has value $m – but generally I find that Perl's hash functions are ferociously optimised and usefully fast.)

    my %q=map {$s{$_} => $_} keys %s;
    return $q{$m};

(If I did care about retaining all the data I'd do something like map {push @{$q{$s{$_}}},$_} sort keys %s; instead and get a series of arrayrefs in %q.)

Otherwise we just return the specified error value.

  } else {
    return -1;

Raku works much the same way.

sub majority(@list) {
  my %s;
  map {%s{$_}++}, @list;
  my $m=max(values %s);
  if ($m > floor(@list.elems/2)) {
    my %q=map {%s{$_} => $_},keys %s;
    return %q{$m};
  } else {
    return -1;

And for an additional challenge I thought I'd give it a go in Python. (I've been meaning to pick this up for a while, and the way I learn languages best is to use them.) Some of the inspirations for Perl6/Raku are obvious. I'll give this first one in full.

#! /usr/bin/python3

import unittest

def majority(list):

We don't have autovivification of hash ("dict") elements, so I have to setdefault each one before incrementing it. (I'm going to miss that.)

    for x in list:
        s[x] += 1
    if m > int(len(list)/2):

The hash inversion works basically as before, even if I find map a more elegant way of doing it.

        for x in s.keys():
        return q[m]
        return -1

And here's some testing infrastructure, which works basically like the test frameworks in Perl. (I want to pass the whole list as one argument, hence the double parentheses.)

class TestMajority(unittest.TestCase):

    def test_ex1(self):
        self.assertEqual(majority((1, 2, 2, 3, 2, 4, 2)),2,'example 1')

    def test_ex2(self):
        self.assertEqual(majority((1, 3, 1, 2, 4, 5)),-1,'example 2')


2. FNR Character

You are given a string $S.

Write a script to print the series of first non-repeating character (left -> right) for the given string. Print # if none found.

(The examples make it clear that what's actually wanted is the last character that's only occurred once so far.)

sub fnr {
  my $in=shift;
  my %s;
  my @s;
  my @o;

So this is a bit fiddly. @s is the list of characters that's occurred so far which haven't repeated. %s is the mapping of those characters to their counts. And @o is the output list.

  foreach my $c (split '',$in) {
    push @s,$c;

Here we strip @s of anything which has occurred more than once.

    @s=grep {$s{$_}<2} @s;

And the last character in @s, if any, is the one we want. (If it were really left-to-right, this would be $s[0].)

    if (@s) {
      push @o,$s[-1];
    } else {
      push @o,'#';
  return join('',@o);

Raku is similar, though we can't use array index -1 any more.

sub fnr($in) {
  my %s;
  my @s;
  my @o;
  for $in.comb -> $c {
    push @s,$c;
    @s=grep {%s{$_} < 2},@s;
    if (@s) {
      push @o,@s[@s.end];
    } else {
      push @o,'#';
  return join('',@o);

And in Python. ls is the list-s, equivalent of @s above, while s is the %s. in is a reserved word.

def fnr(i):

Python strings think they're arrays. How quaint! But it makes it handy for splitting. (The actual split function can't do this.)

    for c in list(i):
        s[c] += 1

But this list comprehension is quite fun, a sort of composition of map and grep from perl. I can see I'll use this a lot.

        ls=[x for x in ls if s[x]<2]
        if len(ls)>0:

What a strange syntax for "join the elements of this list with the given separator". o.join('') (which is what Raku does) makes rather more sense to me. Hey ho.

    return ''.join(o)

So far Python feels more verbose than Perl, but not in a bad way. We'll see.

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