# [LeetCode] 212. Word Search II

## Problem

Given a 2D board and a list of words from the dictionary, find all words in the board.

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

Example:

``````Input:
words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]

Output: ["eat","oath"]``````

Note:
You may assume that all inputs are consist of lowercase letters a-z.

## Solution

``````class Solution {
class TrieNode {
TrieNode[] children = new TrieNode[26];
String word;
}

private TrieNode buildTrie(String[] words) {
TrieNode root = new TrieNode();
for (String word: words) {
TrieNode cur = root;
for (char ch: word.toCharArray()) {
int i = ch-'a';
if (cur.children[i] == null) cur.children[i] = new TrieNode();
cur = cur.children[i];
}
cur.word = word;
}
return root;
}

public List<String> findWords(char[][] board, String[] words) {
List<String> res = new ArrayList<>();
TrieNode root = buildTrie(words);
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
dfs(board, i, j, root, res);
}
}
return res;
}

private void dfs(char[][] board, int i, int j, TrieNode root, List<String> res) {
char ch = board[i][j];
if (ch == '#' || root.children[ch-'a'] == null) return;

root = root.children[ch-'a'];
if (root.word != null) { //found one
root.word = null; //de-dup
}

board[i][j] = '#';
if (i > 0) dfs(board, i-1, j, root, res);
if (j > 0) dfs(board, i, j-1, root, res);
if (i < board.length-1) dfs(board, i+1, j, root, res);
if (j < board[0].length-1) dfs(board, i, j+1, root, res);
board[i][j] = ch;
}
}``````