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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.

Solutions

Solution 1

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class Trie:
    def __init__(self):
        self.children: List[Trie | None] = [None] * 26
        self.ref: int = -1

    def insert(self, w: str, ref: int):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.ref = ref


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        def dfs(node: Trie, i: int, j: int):
            idx = ord(board[i][j]) - ord('a')
            if node.children[idx] is None:
                return
            node = node.children[idx]
            if node.ref >= 0:
                ans.append(words[node.ref])
                node.ref = -1
            c = board[i][j]
            board[i][j] = '#'
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and board[x][y] != '#':
                    dfs(node, x, y)
            board[i][j] = c

        tree = Trie()
        for i, w in enumerate(words):
            tree.insert(w, i)
        m, n = len(board), len(board[0])
        ans = []
        for i in range(m):
            for j in range(n):
                dfs(tree, i, j)
        return ans

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class Trie {
    Trie[] children = new Trie[26];
    int ref = -1;

    public void insert(String w, int ref) {
        Trie node = this;
        for (int i = 0; i < w.length(); ++i) {
            int j = w.charAt(i) - 'a';
            if (node.children[j] == null) {
                node.children[j] = new Trie();
            }
            node = node.children[j];
        }
        node.ref = ref;
    }
}

class Solution {
    private char[][] board;
    private String[] words;
    private List<String> ans = new ArrayList<>();

    public List<String> findWords(char[][] board, String[] words) {
        this.board = board;
        this.words = words;
        Trie tree = new Trie();
        for (int i = 0; i < words.length; ++i) {
            tree.insert(words[i], i);
        }
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                dfs(tree, i, j);
            }
        }
        return ans;
    }

    private void dfs(Trie node, int i