212. Word Search II

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.

For example, Given words = ["oath","pea","eat","rain"] and board =

[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Return ["eat","oath"].

Note:

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

this question requires usage of Trie, 208. Implement Trie (Prefix Tree) and 79 Word Search

public class Solution {
class TrieNode {
    // Initialize your data structure here.
    TrieNode[] children = new TrieNode[26];
    String word = "";
    public TrieNode() {

    }
}

class Trie {
    private TrieNode root;

    public Trie() {
        root = new TrieNode();
    }

    // Inserts a word into the trie.
    public void insert(String word) {
        TrieNode node = root;
        for(char ch : word.toCharArray()){
            if(node.children[ch - 'a'] == null){
                node.children[ch-'a'] = new TrieNode();
            }   
            node = node.children[ch - 'a'];
        }

        node.word = word;
    }

    // Returns if the word is in the trie.
    public boolean search(String word) {
        TrieNode node = root;
        for(char ch : word.toCharArray()){
            if(node.children[ch - 'a'] == null) return false;
            node = node.children[ch - 'a'];
        }
        return node.word.equals(word);
    }

    // Returns if there is any word in the trie
    // that starts with the given prefix.
    public boolean startsWith(String prefix) {
        TrieNode node = root;
        for(char ch : prefix.toCharArray()){
            if(node.children[ch - 'a'] == null) return false;
            node = node.children[ch - 'a'];
        }
        return true;
    }
}

    private Trie dict = new Trie();
    private Set<String> res = new HashSet<>();
    public List<String> findWords(char[][] board, String[] words) {
        if(board == null || board.length == 0 || board[0].length == 0) return new ArrayList<String>(res);
        for(String s : words){
            dict.insert(s);
        }
        boolean[][] visited = new boolean[board.length][board[0].length];
        for(int i =0 ; i< board.length; i++){
            for(int j =0; j< board[0].length; j++){
                dfs(board, visited, i, j, "");
            }
        }

        return new ArrayList<String>(res);
    }

    private void dfs(char[][] board, boolean[][] visited, int x, int y, String current){
        if(x < 0 || y < 0 || x >= board.length || y >= board[0].length || visited[x][y]) return;

        current += board[x][y];
        if(!dict.startsWith(current)) return;
        if(dict.search(current)){
            res.add(current);
        }
        visited[x][y] = true;

        dfs(board, visited, x+1, y, current);
        dfs(board, visited, x-1, y, current);
        dfs(board, visited, x, y+1, current);
        dfs(board, visited, x, y-1, current);

        visited[x][y] = false;
    }

}