Question

Any idea whats a simple and fast way to get all words placed on a Scrabble board while the board is represented by 2D Array of chars?

Thanks in advance

Answer 1

This is similar to Artsiom’s answer, but covers for the fact that a Scrabble board will have spaces in between words.

So assuming your “2D Array of chars” looks like this:

board = [['s','t','a','c','k',' ',' ',' '],
         ['p',' ',' ','a',' ','c',' ',' '],
         ['o','v','e','r','f','l','o','w'],
         ['o',' ','a','t',' ','a',' ','a'],
         ['n','o','t',' ',' ','m','a','t'],
         [' ',' ','e',' ',' ',' ',' ','e'],
         [' ',' ','r',' ',' ',' ',' ','r'],
         [' ','e','y','e','s',' ',' ',' ']]

You can do the following:

import itertools
rows = (''.join(row) for row in board)
columns = (''.join(column) for column in zip(*board))
words = [word for line in itertools.chain(rows,columns) for word in line.split() if len(word) > 1]

Which gives:

['stack', 'overflow', 'at', 'not', 'mat', 'eyes', 'spoon', 'eatery', 'cart', 'clam', 'water']

What we’re doing is converting each row and column of characters in to strings like 'not mat' and then using str.split() to throw away the spaces to give us a list of words, throwing away anything that’s one letter long.

Using itertools.chain() just allows us to loop through the rows and columns in a single list comprehension.