This is a Leetcode problem -

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of a sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.


  • All inputs are consist of lowercase letters a-z.
  • The values of words are distinct.

Here is my solution to this challenge -

class Solution:
    def __init__(self, board, words):
        self.board = board
        self.words = words

    def find_words(self, board, words):

        root = {}
        for word in words:
            node = root
            for c in word:
                node = node.setdefault(c, {})
            node[None] = True
        board = {i + 1j * j: c
                 for i, row in enumerate(board)
                 for j, c in enumerate(row)}

        found = []
        def search(node, z, word):
            if node.pop(None, None):
            c = board.get(z)
            if c in node:
                board[z] = None
                for k in range(4):
                    search(node[c], z + 1j ** k, word + c)
                board[z] = c
        for z in board:
            search(root, z, '')

        return found

Program explanation - I first build a tree of words with root root and also represent the board a different way, namely as a one-dimensional dictionary where the keys are complex numbers representing the row/column indexes. That makes further work with it easier. Looping over all board positions is just for z in board, the four neighbors of a board position z are just z + 1j ** k (for k in 0 to 3), and I don't need to check borders because board.get just returns None if I request an invalid position.

After this preparation, I just take the tree and recursively dive with it into each board position. Similar to how you'd search a single word, but with the tree instead.

Here is an example input/output -

output = Solution([
], ["oath","pea","eat","rain"])

], ["oath","pea","eat","rain"]))

>>> ['oath', 'eat']

So, I would like to know whether I could make this program shorter and more efficient.

Any help would be highly appreciated.


1 Answer 1


Awkward API

It seems strange to initialise a Solution object with various info that we provide again in the call of the method find_words.

Actually, looking for places where self is used makes it pretty obvious: the instance is never used.

It is a good occasion to watch the Stop Writing Classes talk from Jack Diederich.

To be continued ?


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