0212. Word Search II

https://leetcode.com/problems/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.

ac

class Solution {
    public List<String> findWords(char[][] board, String[] words) {
        List<String> res = new ArrayList<String>();
        // edge cases
        if (board.length == 0 || board[0].length == 0 || words.length == 0) {
            return res;
        }

        // build trie
        TrieNode root = new TrieNode();
        root.build(words);

        for (int r = 0; r < board.length; r++) {
            for (int c = 0; c < board[0].length; c++) {
                dfs(board, r, c, root, res);
            }
        }

        return res;
    }

    private void dfs(char[][] board, int r, int c, TrieNode node, List<String> res) {
        // out of boundary?
        if (r < 0 || c < 0 || r >= board.length || c >= board[0].length) return;

        // visited in this word? current char not in trie? return
        int pos = board[r][c] - 'a';
        if (board[r][c] == '#' || node.next[pos] == null) return;

        // continue, mark visited
        char curr = board[r][c];
        board[r][c] = '#';

        // find word
        if (node.next[pos].word != null) {
            res.add(node.next[pos].word);
            node.next[pos].word = null;
        }

        // adjacent chars
        dfs(board, r+1, c, node.next[pos], res);
        dfs(board, r-1, c, node.next[pos], res);
        dfs(board, r, c+1, node.next[pos], res);
        dfs(board, r, c-1, node.next[pos], res);

        // resume visited mark
        board[r][c] = curr;
    }

}

class TrieNode {
    public TrieNode[] next;
    String word;

    public TrieNode() {
        next = new TrieNode[26];
    }

    public void insert(String s, int index) {
        if (index == s.length()) {
            this.word = s;
            return;
        }
        int i = s.charAt(index) - 'a';
        if (next[i] == null) next[i] = new TrieNode();
        next[i].insert(s, index+1);
    }

    public void build(String[] words){
        for (String w : words) {
            insert(w, 0);
        }
    }
}

// build a trie, node with a string
// walk board, current char in trie? continue : return;
//board[r][c] - 'a'