212-word-search-ii¶

Description¶

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.

Example 1:

Input: board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
Output: ["eat","oath"]

Example 2:

Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []

Constraints:

m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] is a lowercase English letter.
1 <= words.length <= 3 * 10⁴
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.
All the strings of words are unique.

Solution(Python)¶

class Trie:
    def __init__(self):
        self.root = {}

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


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = Trie()
        for word in words:
            trie.insert(word)

        m, n = len(board), len(board[0])
        dirs = [[-1, 0], [1, 0], [0, 1], [0, -1]]
        res = []

        def dfs(root, i, j):
            c = board[i][j]
            if c not in root:
                return
            child = root[c]
            word = child.pop('#', None)
            if word:
                res.append(word)
            if not child:
                root.pop(c)

            board[i][j] = '#'
            for dx, dy in dirs:
                x = i + dx
                y = j + dy
                if 0 <= x < m and 0 <= y < n:
                    dfs(child, x, y)
            board[i][j] = c

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

        return res