Word Search II LeetCode Solution

Problem – Word Search II

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: []


  • 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.

Word Search II LeetCode Solution in Java

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;

public void dfs(char[][] board, int i, int j, TrieNode p, List<String> res) {
    char c = board[i][j];
    if (c == '#' || p.next[c - 'a'] == null) return;
    p = p.next[c - 'a'];
    if (p.word != null) {   // found one
        p.word = null;     // de-duplicate

    board[i][j] = '#';
    if (i > 0) dfs(board, i - 1, j ,p, res); 
    if (j > 0) dfs(board, i, j - 1, p, res);
    if (i < board.length - 1) dfs(board, i + 1, j, p, res); 
    if (j < board[0].length - 1) dfs(board, i, j + 1, p, res); 
    board[i][j] = c;

public TrieNode buildTrie(String[] words) {
    TrieNode root = new TrieNode();
    for (String w : words) {
        TrieNode p = root;
        for (char c : w.toCharArray()) {
            int i = c - 'a';
            if (p.next[i] == null) p.next[i] = new TrieNode();
            p = p.next[i];
       p.word = w;
    return root;

class TrieNode {
    TrieNode[] next = new TrieNode[26];
    String word;

Word Search II LeetCode Solution in Python

class TrieNode():
    def __init__(self):
        self.children = collections.defaultdict(TrieNode)
        self.isWord = False
class Trie():
    def __init__(self):
        self.root = TrieNode()
    def insert(self, word):
        node = self.root
        for w in word:
            node = node.children[w]
        node.isWord = True
    def search(self, word):
        node = self.root
        for w in word:
            node = node.children.get(w)
            if not node:
                return False
        return node.isWord
class Solution(object):
    def findWords(self, board, words):
        res = []
        trie = Trie()
        node = trie.root
        for w in words:
        for i in xrange(len(board)):
            for j in xrange(len(board[0])):
                self.dfs(board, node, i, j, "", res)
        return res
    def dfs(self, board, node, i, j, path, res):
        if node.isWord:
            node.isWord = False
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]):
        tmp = board[i][j]
        node = node.children.get(tmp)
        if not node:
        board[i][j] = "#"
        self.dfs(board, node, i+1, j, path+tmp, res)
        self.dfs(board, node, i-1, j, path+tmp, res)
        self.dfs(board, node, i, j-1, path+tmp, res)
        self.dfs(board, node, i, j+1, path+tmp, res)
        board[i][j] = tmp

Word Search II LeetCode Solution in C++

class TrieNode{
    bool is_end;
    vector<TrieNode*> children;
        children=vector<TrieNode*>(26, NULL);

class Trie{
    TrieNode* getRoot(){return root;}
    Trie(vector<string>& words){
        root=new TrieNode();
        for(int i=0; i<words.size(); ++i)
    void addWord(const string& word){
        TrieNode* cur=root;
        for(int i=0; i<word.size(); ++i){
            int index=word[i]-'a';
               cur->children[index]=new TrieNode();
    TrieNode* root;

class Solution {
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        Trie* trie = new Trie(words);
        TrieNode* root=trie->getRoot();
        set<string> result_set;
        for(int x=0; x<board.size(); ++x)
            for(int y=0; y<board[0].size(); ++y)
                findWords(board, x, y, root, "", result_set);
        vector<string> result;
        for(auto it:result_set)    result.push_back(it);
        return result;        
    void findWords(vector<vector<char>>& board, int x, int y, TrieNode* root, string word, set<string>& result){
        if(x<0||x>=board.size()||y<0||y>=board[0].size() || board[x][y]==' ') return;
        if(root->children[board[x][y]-'a'] != NULL){
            if(root->is_end) result.insert(word);
            char c=board[x][y];
            board[x][y]=' ';
            findWords(board, x+1, y, root, word, result);
            findWords(board, x-1, y, root, word, result);
            findWords(board, x, y+1, root, word, result);
            findWords(board, x, y-1, root, word, result);
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