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Java Java Objects Creating the MVP Counting Scrabble Tiles

I just can't solve this problem, give me some hints.

Help me to get through this problem.

ScrabblePlayer.java
public class ScrabblePlayer {
  // A String representing all of the tiles that this player has
  private String tiles;

  public ScrabblePlayer() {
    tiles = "";
  }

  public String getTiles() {
    return tiles;
  }

  public void addTile(char tile) {
    tiles += tile;
  }

  public boolean hasTile(char tile) {
    return tiles.indexOf(tile) != -1;
  }
  public int getCountOfLetter(char letter) {
    int c = 0;
    for(int i=0;i<tiles.length;i++){
      if(tiles.indexOf(letter) != -1)
        c++;
    }
    return c;
  }
}
Example.java
// This code is here for example purposes only
public class Example {

  public static void main(String[] args) {
    ScrabblePlayer player1 = new ScrabblePlayer();
    player1.addTile('d');
    player1.addTile('d');
    player1.addTile('p');
    player1.addTile('e');
    player1.addTile('l');
    player1.addTile('u');

    ScrabblePlayer player2 = new ScrabblePlayer();
    player2.addTile('z');
    player2.addTile('z');
    player2.addTile('y');
    player2.addTile('f');
    player2.addTile('u');
    player2.addTile('z');

    int count = 0;
    // This would set count to 1 because player1 has 1 'p' tile in her collection of tiles
    count = player1.getCountOfLetter('p');
    // This would set count to 2 because player1 has 2 'd'' tiles in her collection of tiles
    count = player1.getCountOfLetter('d');
    // This would set 0, because there isn't an 'a' tile in player1's tiles
    count = player1.getCountOfLetter('a');

    // This will return 3 because player2 has 3 'z' tiles in his collection of tiles
    count = player2.getCountOfLetter('z');
    // This will return 1 because player2 has 1 'f' tiles in his collection of tiles
    count = player2.getCountOfLetter('f');
  }
}

1 Answer

Steven Parker
Steven Parker
221,070 Points

Using "indexOf" was great for "hasTile", but it's not as useful for counting items because it will always return the same result no matter how many matching tiles exist. A better strategy was mentioned in the instructions: "use an equality check, and then increment a counter if the tile and letter match".

Bonus hint: even though the instructions don't mention it, this challenge specifically wants you to use an enhanced for loop. The "toCharArray" method might be useful for converting the string into a suitable collection.