Question

Formatted question description: https://leetcode.ca/all/212.html

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"].


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.


@tag-trie

Algorithm

The solution to problem 79. Word Search can be reused here. Just search for each word in words and check whether it exists, and add the words that exist to the result list.

There also exist other solutions like Trie + DFS.

Code

Trie

import java.util.ArrayList;
import java.util.List;

public class Word_Search_II {

    public static void main(String[] args) {
        Word_Search_II out = new Word_Search_II();
        Solution s = out.new Solution();

        System.out.println(
            s.findWords(
                new char[][]{
                    {'o','a','a','n'},
                    {'e','t','a','e'},
                    {'i','h','k','r'},
                    {'i','f','l','v'} },
                new String[]{"oath","pea","eat","rain"}
        ));

        System.out.println(
            s.findWords(
                new char[][]{
                    {'a', 'b'},
                    {'a', 'a'} },
                new String[]{"aba","baa","bab","aaab","aaa","aaaa","aaba"}
            )); // output: ["aaa","aaab","aaba","aba","baa"]
    }


    /*

test case: huge data

board:
    [["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"],["a","a","a","a","a","a","a","a","a","a","a","a"]]

words:
    ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

     */

    class TrieNode {
        public TrieNode[] children;
        public String word;

        public TrieNode() {
            children = new TrieNode[26];
            word = null;
            for (int i = 0; i < 26; i++) {
                children[i] = null;
            }
        }
    }

    class Trie {
        TrieNode root;
        /** Initialize your data structure here. */
        public Trie() {
            root = new TrieNode();
        }

        /** Inserts a word into the trie. */
        public void insert(String word) {
            char[] array = word.toCharArray();
            TrieNode cur = root;
            for (int i = 0; i < array.length; i++) {
                if (cur.children[array[i] - 'a'] == null) {
                    cur.children[array[i] - 'a'] = new TrieNode();
                }
                cur = cur.children[array[i] - 'a'];
            }
            cur.word = word;
        }
    };

    class Solution {
        public List<String> findWords(char[][] board, String[] words) {
            Trie trie = new Trie();

            List<String> res = new ArrayList<>();
            for (String word : words) {
                trie.insert(word);
            }
            int rows = board.length;
            if (rows == 0) {
                return res;
            }
            int cols = board[0].length;

            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < cols; j++) {
                    dfs(board, i, j, trie.root, res);
                }
            }
            return res;
        }

        private void dfs(char[][] board, int row, int col, TrieNode node, List<String> res) {
            if (row < 0 || row >= board.length || col < 0 || col >= board[0].length) {
                return;
            }
            char cur = board[row][col];
            
            if (cur == '$' || node.children[cur - 'a'] == null) {
                return;
            }
            node = node.children[cur - 'a'];

            if (node.word != null) {
                res.add(node.word);
                node.word = null;
            }
            char temp = board[row][col];

            board[row][col] = '$';
            dfs(board, row - 1, col, node, res);
            dfs(board, row + 1, col, node, res);
            dfs(board, row, col - 1, node, res);
            dfs(board, row, col + 1, node, res);
            board[row][col] = temp;
        }
    }
}

// OJ: https://leetcode.com/problems/word-search-ii/
// Time: O(WL + MN * 4^L) where M, N is the size of board, W is the size of words and L is the average length of word
// Space: O(WL)
struct TrieNode {
    TrieNode *next[26] = {0};
    bool word = false;
};
void addWord(TrieNode *node, string &w) {
    for (char c : w) {
        if (node->next[c - 'a'] == NULL) node->next[c - 'a'] = new TrieNode();
        node = node->next[c - 'a'];
    }
    node->word = true;
}
class Solution {
    int M, N, dirs[4][2] = { {0,1}, {0,-1}, {1,0}, {-1,0} };
    vector<string> ans;
    string path;
    void dfs(vector<vector<char>> &A, TrieNode *node, int x, int y) {
        char c = A[x][y];
        node = node->next[c - 'a'];
        if (!node) return;
        path.push_back(c);
        A[x][y] = 0;
        if (node->word) {
            ans.push_back(path);
            node->word = false; // prevent adding the same word again.
        }
        for (auto &dir : dirs) {
            int a = x + dir[0], b = y + dir[1];
            if (a >= M || a < 0 || b >= N || b < 0 || A[a][b] == 0) continue;
            dfs(A, node, a, b);
        }
        A[x][y] = c;
        path.pop_back();
    }
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        if (board.empty() || board[0].empty()) return {};
        M = board.size(), N = board[0].size();
        TrieNode root;
        for (auto &w : words) addWord(&root, w);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) dfs(board, &root, i, j);
        }
        return ans;
    }
};

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.w = ''

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.w = w


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        def dfs(node, i, j):
            idx = ord(board[i][j]) - ord('a')
            if node.children[idx] is None:
                return
            node = node.children[idx]
            if node.w:
                ans.add(node.w)
            c = board[i][j]
            board[i][j] = '0'
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and board[x][y] != '0':
                    dfs(node, x, y)
            board[i][y] = c

        trie = Trie()
        for w in words:
            trie.insert(w)
        ans = set()
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                dfs(trie, i, j)
        return list(ans)

############

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.w = ''

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.w = w


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        def dfs(node, i, j):
            idx = ord(board[i][j]) - ord('a')
            if node.children[idx] is None:
                return
            node = node.children[idx]
            if node.w:
                ans.add(node.w)
            c = board[i][j]
            board[i][j] = '0'
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and board[x][y] != '0':
                    dfs(node, x, y)
            board[i][y] = c

        trie = Trie()
        for w in words:
            trie.insert(w)
        ans = set()
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                dfs(trie, i, j)
        return list(ans)

################3

class TrieNode:
  def __init__(self, char):
    self.neighbours = {}
    self.isWord = False


class Trie:
  def __init__(self):
    self.root = TrieNode("-")

  def addWord(self, word):
    root = self.root
    for i in range(0, len(word)):
      c = word[i]
      if c in root.neighbours:
        root = root.neighbours[c]
      else:
        newnode = TrieNode(c)
        root.neighbours[c] = newnode
        root = root.neighbours[c]
    root.isWord = True


class Solution:
  # @param board, a list of lists of 1 length string
  # @param words: A list of string
  # @return: A list of string
  def findWords(self, board, words):
    # write your code here
    trie = Trie()
    res = []
    visited = [[0] * len(board[0]) for i in range(0, len(board))]
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

    def dfs(i, j, board, visited, res, root, path):
      if not root:
        return

      if root.isWord:
        res.append(path)

      for direction in directions:
        ni, nj = i + direction[0], j + direction[1]
        if 0 <= ni < len(board) and 0 <= nj < len(board[0]):
          c = board[ni][nj]
          if visited[ni][nj] == 0:
            visited[ni][nj] = 1
            dfs(ni, nj, board, visited, res, root.neighbours.get(c, None), path + c)
            visited[ni][nj] = 0

    for word in words:
      trie.addWord(word)
    root = trie.root
    for i in range(0, len(board)):
      for j in range(0, len(board[0])):
        c = board[i][j]
        visited[i][j] = 1
        dfs(i, j, board, visited, res, root.neighbours.get(c, None), c)
        visited[i][j] = 0
    return list(set(res))

Java

class Solution {
    public List<String> findWords(char[][] board, String[] words) {
        List<String> wordsInBoard = new ArrayList<String>();
        for (String word : words) {
            if (exist(board, word))
                wordsInBoard.add(word);
        }
        return wordsInBoard;
    }

    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || board[0].length == 0 || word == null || word.length() == 0)
            return false;
        int rows = board.length, columns = board[0].length;
        boolean[][] visited = new boolean[rows][columns];
        char startLetter = word.charAt(0);
        List<int[]> startPositions = new ArrayList<int[]>();
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (board[i][j] == startLetter)
                    startPositions.add(new int[]{i, j});
            }
        }
        for (int[] startPosition : startPositions) {
            boolean exist = depthFirstSearch(board, word, startPosition, visited);
            if (exist)
                return true;
        }
        return false;
    }

    public boolean depthFirstSearch(char[][] board, String word, int[] startPosition, boolean[][] visited) {
        return depthFirstSearch(board, word, startPosition, visited, 0);
    }

    public boolean depthFirstSearch(char[][] board, String word, int[] startPosition, boolean[][] visited, int startIndex) {
        int rows = board.length, columns = board[0].length;
        int length = word.length();
        int startRow = startPosition[0], startColumn = startPosition[1];
        if (startIndex == length - 1)
            return board[startRow][startColumn] == word.charAt(startIndex);
        if (board[startRow][startColumn] != word.charAt(startIndex))
            return false;
        visited[startRow][startColumn] = true;
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        for (int[] direction : directions) {
            int newRow = startRow + direction[0], newColumn = startColumn + direction[1];
            if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && !visited[newRow][newColumn]) {
                int[] newPosition = {newRow, newColumn};
                int newIndex = startIndex + 1;
                if (depthFirstSearch(board, word, newPosition, visited, newIndex))
                    return true;
            }
        }
        visited[startRow][startColumn] = false;
        return false;
    }
}

// OJ: https://leetcode.com/problems/word-search-ii/
// Time: O(WL + MN * 4^L) where M, N is the size of board, W is the size of words and L is the average length of word
// Space: O(WL)
struct TrieNode {
    TrieNode *next[26] = {0};
    bool word = false;
};
void addWord(TrieNode *node, string &w) {
    for (char c : w) {
        if (node->next[c - 'a'] == NULL) node->next[c - 'a'] = new TrieNode();
        node = node->next[c - 'a'];
    }
    node->word = true;
}
class Solution {
    int M, N, dirs[4][2] = { {0,1}, {0,-1}, {1,0}, {-1,0} };
    vector<string> ans;
    string path;
    void dfs(vector<vector<char>> &A, TrieNode *node, int x, int y) {
        char c = A[x][y];
        node = node->next[c - 'a'];
        if (!node) return;
        path.push_back(c);
        A[x][y] = 0;
        if (node->word) {
            ans.push_back(path);
            node->word = false; // prevent adding the same word again.
        }
        for (auto &dir : dirs) {
            int a = x + dir[0], b = y + dir[1];
            if (a >= M || a < 0 || b >= N || b < 0 || A[a][b] == 0) continue;
            dfs(A, node, a, b);
        }
        A[x][y] = c;
        path.pop_back();
    }
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        if (board.empty() || board[0].empty()) return {};
        M = board.size(), N = board[0].size();
        TrieNode root;
        for (auto &w : words) addWord(&root, w);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) dfs(board, &root, i, j);
        }
        return ans;
    }
};

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.w = ''

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.w = w


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        def dfs(node, i, j):
            idx = ord(board[i][j]) - ord('a')
            if node.children[idx] is None:
                return
            node = node.children[idx]
            if node.w:
                ans.add(node.w)
            c = board[i][j]
            board[i][j] = '0'
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and board[x][y] != '0':
                    dfs(node, x, y)
            board[i][y] = c

        trie = Trie()
        for w in words:
            trie.insert(w)
        ans = set()
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                dfs(trie, i, j)
        return list(ans)

############

class TrieNode:
  def __init__(self, char):
    self.neighbours = {}
    self.isWord = False


class Trie:
  def __init__(self):
    self.root = TrieNode("-")

  def addWord(self, word):
    root = self.root
    for i in range(0, len(word)):
      c = word[i]
      if c in root.neighbours:
        root = root.neighbours[c]
      else:
        newnode = TrieNode(c)
        root.neighbours[c] = newnode
        root = root.neighbours[c]
    root.isWord = True


class Solution:
  # @param board, a list of lists of 1 length string
  # @param words: A list of string
  # @return: A list of string
  def findWords(self, board, words):
    # write your code here
    trie = Trie()
    res = []
    visited = [[0] * len(board[0]) for i in range(0, len(board))]
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

    def dfs(i, j, board, visited, res, root, path):
      if not root:
        return

      if root.isWord:
        res.append(path)

      for direction in directions:
        ni, nj = i + direction[0], j + direction[1]
        if 0 <= ni < len(board) and 0 <= nj < len(board[0]):
          c = board[ni][nj]
          if visited[ni][nj] == 0:
            visited[ni][nj] = 1
            dfs(ni, nj, board, visited, res, root.neighbours.get(c, None), path + c)
            visited[ni][nj] = 0

    for word in words:
      trie.addWord(word)
    root = trie.root
    for i in range(0, len(board)):
      for j in range(0, len(board[0])):
        c = board[i][j]
        visited[i][j] = 1
        dfs(i, j, board, visited, res, root.neighbours.get(c, None), c)
        visited[i][j] = 0
    return list(set(res))

212 - Word Search II

Question

Algorithm

Code

All Problems

All Solutions

Welcome to Subscribe On Youtube

Question

Algorithm

Code

All Problems

All Solutions