# LeetCode – Word Search II (Java)

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"].

Java Solution 1

Similar to Word Search, this problem can be solved by DFS. However, this solution exceeds time limit.

```public List<String> findWords(char[][] board, String[] words) { ArrayList<String> result = new ArrayList<String>();   int m = board.length; int n = board[0].length;   for (String word : words) { boolean flag = false; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { char[][] newBoard = new char[m][n]; for (int x = 0; x < m; x++) for (int y = 0; y < n; y++) newBoard[x][y] = board[x][y];   if (dfs(newBoard, word, i, j, 0)) { flag = true; } } } if (flag) { result.add(word); } }   return result; }   public boolean dfs(char[][] board, String word, int i, int j, int k) { int m = board.length; int n = board[0].length;   if (i < 0 || j < 0 || i >= m || j >= n || k > word.length() - 1) { return false; }   if (board[i][j] == word.charAt(k)) { char temp = board[i][j]; board[i][j] = '#';   if (k == word.length() - 1) { return true; } else if (dfs(board, word, i - 1, j, k + 1) || dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i, j - 1, k + 1) || dfs(board, word, i, j + 1, k + 1)) { board[i][j] = temp; return true; }   } else { return false; }   return false; }```

Java Solution 2 - Trie

If the current candidate does not exist in all words' prefix, we can stop backtracking immediately. This can be done by using a trie structure.

```public class Solution { Set<String> result = new HashSet<String>();   public List<String> findWords(char[][] board, String[] words) { //HashSet<String> result = new HashSet<String>();   Trie trie = new Trie(); for(String word: words){ trie.insert(word); }   int m=board.length; int n=board[0].length;   boolean[][] visited = new boolean[m][n];   for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ dfs(board, visited, "", i, j, trie); } }   return new ArrayList<String>(result); }   public void dfs(char[][] board, boolean[][] visited, String str, int i, int j, Trie trie){ int m=board.length; int n=board[0].length;   if(i<0 || j<0||i>=m||j>=n){ return; }   if(visited[i][j]) return;   str = str + board[i][j];   if(!trie.startsWith(str)) return;   if(trie.search(str)){ result.add(str); }   visited[i][j]=true; dfs(board, visited, str, i-1, j, trie); dfs(board, visited, str, i+1, j, trie); dfs(board, visited, str, i, j-1, trie); dfs(board, visited, str, i, j+1, trie); visited[i][j]=false; } }```
```//Trie Node class TrieNode{ public TrieNode[] children = new TrieNode[26]; public String item = ""; }   //Trie class Trie{ public TrieNode root = new TrieNode();   public void insert(String word){ TrieNode node = root; for(char c: word.toCharArray()){ if(node.children[c-'a']==null){ node.children[c-'a']= new TrieNode(); } node = node.children[c-'a']; } node.item = word; }   public boolean search(String word){ TrieNode node = root; for(char c: word.toCharArray()){ if(node.children[c-'a']==null) return false; node = node.children[c-'a']; } if(node.item.equals(word)){ return true; }else{ return false; } }   public boolean startsWith(String prefix){ TrieNode node = root; for(char c: prefix.toCharArray()){ if(node.children[c-'a']==null) return false; node = node.children[c-'a']; } return true; } }```
Category >> Algorithms >> Interview
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• WanderingTycoon

Does anyone know why we can’t use a StringBuilder to improve efficiency for this case?