Solving with Ruby, Perl, Java and Python AtCoder diverta 2019 Programming Contest C String Manipulation

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This theme

AtCoder diverta 2019 Programming Contest C - AB Substrings Difficulty: 911

This theme, string manipulation Ruby Each given string is counted by BXA`` BX XA. Of these, BXA becomes BXA even if BXAs are combined, so process this first to reduce the number of BXAs. Similarly, even if BXA and BX are combined, it is BX, and even if XA and BXA are combined, it is XA, so use that to count.

ruby.rb

n = gets.to_i
w = x = y = z = 0
n.times do |i|
  s = gets.chomp
  w += s.scan("AB").size
  if s[0] == "B"
    if s[-1] == "A"
      x += 1
    else
      y += 1
    end
  elsif s[-1] == "A"
    z += 1
  end
end
if x > 1
  w += x - 1
  x = 1
end
if [x, y, z].max == y
  w += x
  w += z
elsif [x, y, z].max == z
  w += x
  w += y
end
puts w

scan.rb

  w += s.scan("AB").size

For * Ruby *, s.scan ("AB"). Size counts the number of ʻAB`.

string.rb

    if s[-1] == "A"

It's amazing that you can access the last character with [-1]. Python

python.py

import re

n = int(input())
w = x = y = z = 0
for i in range(n):
    s = input()
    t = re.sub(r'AB', "X", s)
    w += len(s) - len(t)
    if s[0] == "B":
        if s[-1] == "A":
            x += 1
        else:
            y += 1
    elif s[-1] == "A":
        z += 1
if x > 1:
    w += x - 1
    x = 1
if max(x, y, z) == y:
    w += x
    w += z
elif max(x, y, z) == z:
    w += x
    w += y
print(w)

len.py

    t = re.sub(r'AB', "X", s)
    w += len(s) - len(t)

I'm not familiar with * Python * functions, so I'm counting the number of ʻAB` from the number of characters before and after replacement. Perl

perl.pl

use List::Util qw/max/;

chomp (my $n = <STDIN>);
my ($w, $x, $y, $z);
for my $i (1..$n) {
  chomp (my $s = <STDIN>);
  $w++ while $s =~ /AB/g;
  if (substr($s, 0, 1) eq "B") {
    if (substr($s, -1, 1) eq "A") {
      $x++;
    } else {
      $y++;
    }
  } elsif (substr($s, -1, 1) eq "A") {
    $z++;
  }
}
if ($x > 1) {
  $w += $x - 1;
  $x = 1;
}
if (max($x, $y, $z) == $y) {
  $w += $x;
  $w += $z;
} elsif (max($x, $y, $z) == $z) {
  $w += $x;
  $w += $y;
}
print "$w\n";

reg.pl

  $w++ while $s =~ /AB/g;

For * Perl *, there is a idiom. After all, Perl is great for character processing. Java

java.java

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int w = 0, x = 0, y = 0, z = 0;
        for (int i = 0; i < n; i++) {
            String s = sc.next();
            String t = s.replaceAll("AB", "X");
            w += s.length() - t.length();
            if (s.substring(0, 1).equals("B")) {
                if (s.substring(s.length() - 1, s.length()).equals("A")) {
                    x++;
                } else {
                    y++;
                }
            } else if (s.substring(s.length() - 1, s.length()).equals("A")) {
                z++;
            }
        }
        sc.close();
        if (x > 1) {
            w += x - 1;
            x = 1;
        }
        if (Math.max(x, Math.max(y, z)) == y) {
            w += x;
            w += z;
        } else if (Math.max(x, Math.max(y, z)) == z) {
            w += x;
            w += y;
        }
        System.out.println(w);
    }
}

replaceAll.java

            String t = s.replaceAll("AB", "X");
            w += s.length() - t.length();

In the case of * Java *, the number of ʻAB` is counted from the number of characters before and after replacement as in Python.

Summary

• Solved diverta 2019 Programming Contest C
• Become familiar with Ruby
• Become familiar with Python
• Become familiar with Perl
• Become familiar with Java

Referenced site

• Count the number of occurrences of a character string (two or more characters) in Ruby *
• Perl table technique collection: Get the number of matches *