Leetcode 212. Word Search II

Problem

Given an m x n board of characters and a list of strings words, return all words on the board.

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

Algorithm

Use Trie for save and search word, run dfs find the word in the board.

Code

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        
        class Trie:
            def __init__(self):
                self.root = {}

            def insert(self, word):
                node = self.root
                for c in word:
                    if c not in node:
                        node[c] = {}
                    node = node[c]
                node['leaf'] = word  

        trie = Trie()
        for word in words:
            trie.insert(word)

        m, n = len(board), len(board[0])
        result = []

        def dfs(x, y, node):
            c = board[x][y]
            if c not in node:
                return
            next_node = node[c]
            word = next_node.get('leaf')
            if word:
                result.append(word)
                next_node['leaf'] = None # need remove

            board[x][y] = '#'
            for dx, dy in [(-1,0), (1,0), (0,-1), (0,1)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < m and 0 <= ny < n and board[nx][ny] != '#':
                    dfs(nx, ny, next_node)
            board[x][y] = c

        for i in range(m):
            for j in range(n):
                dfs(i, j, trie.root)

        return list(result)