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: []

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.

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
        res.add(p.word);
        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:
            trie.insert(w)
        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:
            res.append(path)
            node.isWord = False
        if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]):
            return 
        tmp = board[i][j]
        node = node.children.get(tmp)
        if not node:
            return 
        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{
public:
    bool is_end;
    vector<TrieNode*> children;
    TrieNode(){
        is_end=false;
        children=vector<TrieNode*>(26, NULL);
    }   
};

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

class Solution {
public:
    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;        
    }
private:
    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){
            word=word+board[x][y];
            root=root->children[board[x][y]-'a']; 
            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);
            board[x][y]=c;        
        }
    }
};
Word Search II LeetCode Solution Review:

In our experience, we suggest you solve this Word Search II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Word Search II LeetCode Solution

Find on LeetCode

Conclusion:

I hope this Word Search II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions

Leave a Reply

Your email address will not be published.