212. Word Search II

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212. Word Search II

Description

Given an m x n boardof 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:

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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:

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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 * 10^4
  • 1 <= words[i].length <= 10
  • words[i] consists of lowercase English letters.
  • All the strings of words are unique.

Hints/Notes

  • 2025/01/02
  • Trie
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    struct TrieNode {
        bool end = false;
        TrieNode* children[26]{};
    };

    static constexpr int DIRs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    unordered_set<string> res;
    string cur;
    int m, n;

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        TrieNode root;
        for (auto& word : words) {
            TrieNode* head = &root;
            for (auto& c : word) {
                if (!head->children[c - 'a']) {
                    head->children[c - 'a'] = new TrieNode();
                }
                head = head->children[c - 'a'];
            }
            head->end = true;
        }

        m = board.size();
        n = board[0].size();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dfs(&root, i, j, board);
            }
        }
        return vector<string>(res.begin(), res.end());
    }

    void dfs(TrieNode* node, int x, int y, vector<vector<char>>& board) {
        char c = board[x][y];
        int index = board[x][y] - 'a';
        if (index < 0) {
            return;
        }
        if (node->children[index] == nullptr) {
            return;
        }
        cur.push_back(c);
        board[x][y] = 'a' - 1;
        node = node->children[index];
        if (node->end) {
            res.insert(cur);
        }
        for (int i = 0; i < 4; i++) {
            int dx = x + DIRs[i][0], dy = y + DIRs[i][1];
            if (dx >= 0 && dx < m && dy >= 0 && dy < n) {
                dfs(node, dx, dy, board);
            }
        }
        cur.pop_back();
        board[x][y] = c;
    }
};