Let's play Scrabble

Problem:


Write a program that, given a word, computes the scrabble score for
that word.

Code:

import java.util.regex.Pattern;
import java.util.Optional;
import java.util.Map;
import java.util.HashMap;

public class Scrabble {
  private static final Pattern WHITE_SPACE = Pattern.compile("\\s+");
  private static final Map SCRABBLE_SCORE_MAP = initMap();
  private final String phrase;
  private int score;

  public Scrabble(String phrase) {
    this.phrase = Optional.ofNullable(phrase).orElse("")
        .replaceAll(WHITE_SPACE.toString(), "").toLowerCase();
    score = calculate();
  }

  public int getScore() {
    return score;
  }

  private int calculate() {
    int score = 0;
    if (phrase.isEmpty()) {
      return score;
    }

    for (char chr : phrase.toCharArray()) {
      for (String key : SCRABBLE_SCORE_MAP.keySet()) {
        if (key.indexOf(chr) != -1) {
          score += SCRABBLE_SCORE_MAP.get(key);
        }
      }
    }
    return score;
  }

  private static Map initMap() {
    Map map = new HashMap<>();
    map.put("aeioulnrst", 1);
    map.put("dg", 2);
    map.put("bcmp", 3);
    map.put("fhvwy", 4);
    map.put("k", 5);
    map.put("jx", 8);
    map.put("qz", 10);
    return map;
  }
}

Test Suite:

```
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;

import java.util.Arrays;
import java.util.Collection;

import static org.junit.Assert.assertEquals;

@RunWith(Parameterized.class)
public class ScrabbleScoreTest {

private String input;
private int expectedOutput;

@Parameterized.Parameters
public static Collection data() {
return Arrays.asList(new Object[][]{
{"", 0},
{" \t\n", 0},
{null, 0},
{"a", 1},
{"f", 4},
{"street", 6},
{"quirky", 22},
{"OXYPHENBUTAZONE", 41},
{"alacrity", 13},
});
}

public S

Some suggestions:

• Make the score map a HashMap so every single rated char is mapped to one integer.

• Do not use Regexes to remove invalid characters. Simply ignore them.

• Move the phrase out of the constructor. Make your calculate() method accept the phrase instead to allow re-use of the same Scrabble object.

So in the end you can simplify the score evaluation a lot:

public int calculate(String input) {
    int score = 0;
    for (char c: input.toCharArray()) {
        score += SCRABBLE_SCORE_MAP.getOrDefault(c, 0);
    }
}

In theory, this should run more performant since there are no Regex transformations, you are not cycling through the whole Map and you are not cycling through the whole String to get the score of a character.

StackExchange Code Review Q#128672, answer score: 4