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Boggle | Set 2 (Using Trie)

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  • Difficulty Level : Expert
  • Last Updated : 24 May, 2022
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Given a dictionary, a method to do a lookup in the dictionary and a M x N board where every cell has one character. Find all possible words that can be formed by a sequence of adjacent characters. Note that we can move to any of 8 adjacent characters, but a word should not have multiple instances of the same cell.
Example: 
 

Input: dictionary[] = {"GEEKS", "FOR", "QUIZ", "GO"};
       boggle[][]   = {{'G', 'I', 'Z'},
                       {'U', 'E', 'K'},
                       {'Q', 'S', 'E'}};

Output: Following words of the dictionary are present
         GEEKS
         QUIZ

Explanation:

Input: dictionary[] = {"GEEKS", "ABCFIHGDE"};
       boggle[][]   = {{'A', 'B', 'C'},
                       {'D', 'E', 'F'},
                       {'G', 'H', 'I'}};
Output: Following words of the dictionary are present
         ABCFIHGDE
Explanation:

 
Recommended Practice

We have discussed a Graph DFS based solution in below post. 
Boggle (Find all possible words in a board of characters) | Set 1
Here we discuss a Trie based solution which is better than DFS based solution. 
Given Dictionary dictionary[] = {“GEEKS”, “FOR”, “QUIZ”, “GO”} 
1. Create an Empty trie and insert all words of given dictionary into trie 
 

After insertion, Trie looks like(leaf nodes are in RED)
                       root
                    /       
                    G   F     Q
                 /  |   |     |
                O   E   O     U
                    |   |     |
                    E    R     I
                    |         |  
                    K         Z 
                    |   
                    S   

2. After that we have pick only those character in boggle[][] which are child of root of Trie 
Let for above we pick ‘G’ boggle[0][0], ‘Q’ boggle[2][0] (they both are present in boggle matrix) 
3. search a word in a trie which start with character that we pick in step 2 
 

1) Create bool visited boolean matrix (Visited[M][N] = false )
2) Call SearchWord() for every cell (i, j) which has one of the
   first characters of dictionary words. In above example,
   we have 'G' and 'Q' as first characters.

SearchWord(Trie *root, i, j, visited[][N])
if root->leaf == true 
   print word 

if we have seen this element first time then make it visited.
   visited[i][j] = true
   do
      traverse all child of current root 
      k goes (0 to 26 ) [there are only 26 Alphabet] 
      add current char and search for next character 

      find next character which is adjacent to boggle[i][j]
      they are 8 adjacent cells of boggle[i][j] (i+1, j+1), 
      (i+1, j) (i-1, j) and so on.
   
   make it unvisited visited[i][j] = false 

Below is the implementation of above idea: 
 

C++




// C++ program for Boggle game
#include <bits/stdc++.h>
using namespace std;
 
// Converts key current character into index
// use only 'A' through 'Z'
#define char_int(c) ((int)c - (int)'A')
 
// Alphabet size
#define SIZE (26)
 
#define M 3
#define N 3
 
// trie Node
struct TrieNode {
    TrieNode* Child[SIZE];
 
    // isLeaf is true if the node represents
    // end of a word
    bool leaf;
};
 
// Returns new trie node (initialized to NULLs)
TrieNode* getNode()
{
    TrieNode* newNode = new TrieNode;
    newNode->leaf = false;
    for (int i = 0; i < SIZE; i++)
        newNode->Child[i] = NULL;
    return newNode;
}
 
// If not present, inserts a key into the trie
// If the key is a prefix of trie node, just
// marks leaf node
void insert(TrieNode* root, char* Key)
{
    int n = strlen(Key);
    TrieNode* pChild = root;
 
    for (int i = 0; i < n; i++) {
        int index = char_int(Key[i]);
 
        if (pChild->Child[index] == NULL)
            pChild->Child[index] = getNode();
 
        pChild = pChild->Child[index];
    }
 
    // make last node as leaf node
    pChild->leaf = true;
}
 
// function to check that current location
// (i and j) is in matrix range
bool isSafe(int i, int j, bool visited[M][N])
{
    return (i >= 0 && i < M && j >= 0 && j < N && !visited[i][j]);
}
 
// A recursive function to print all words present on boggle
void searchWord(TrieNode* root, char boggle[M][N], int i,
                int j, bool visited[][N], string str)
{
    // if we found word in trie / dictionary
    if (root->leaf == true)
        cout << str << endl;
 
    // If both I and j in  range and we visited
    // that element of matrix first time
    if (isSafe(i, j, visited)) {
        // make it visited
        visited[i][j] = true;
 
        // traverse all childs of current root
        for (int K = 0; K < SIZE; K++) {
            if (root->Child[K] != NULL) {
                // current character
                char ch = (char)K + (char)'A';
 
                // Recursively search reaming character of word
                // in trie for all 8 adjacent cells of boggle[i][j]
                if (isSafe(i + 1, j + 1, visited)
                    && boggle[i + 1][j + 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i + 1, j + 1, visited, str + ch);
                if (isSafe(i, j + 1, visited)
                    && boggle[i][j + 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i, j + 1, visited, str + ch);
                if (isSafe(i - 1, j + 1, visited)
                    && boggle[i - 1][j + 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i - 1, j + 1, visited, str + ch);
                if (isSafe(i + 1, j, visited)
                    && boggle[i + 1][j] == ch)
                    searchWord(root->Child[K], boggle,
                               i + 1, j, visited, str + ch);
                if (isSafe(i + 1, j - 1, visited)
                    && boggle[i + 1][j - 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i + 1, j - 1, visited, str + ch);
                if (isSafe(i, j - 1, visited)
                    && boggle[i][j - 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i, j - 1, visited, str + ch);
                if (isSafe(i - 1, j - 1, visited)
                    && boggle[i - 1][j - 1] == ch)
                    searchWord(root->Child[K], boggle,
                               i - 1, j - 1, visited, str + ch);
                if (isSafe(i - 1, j, visited)
                    && boggle[i - 1][j] == ch)
                    searchWord(root->Child[K], boggle,
                               i - 1, j, visited, str + ch);
            }
        }
 
        // make current element unvisited
        visited[i][j] = false;
    }
}
 
// Prints all words present in dictionary.
void findWords(char boggle[M][N], TrieNode* root)
{
    // Mark all characters as not visited
    bool visited[M][N];
    memset(visited, false, sizeof(visited));
 
    TrieNode* pChild = root;
 
    string str = "";
 
    // traverse all matrix elements
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++) {
            // we start searching for word in dictionary
            // if we found a character which is child
            // of Trie root
            if (pChild->Child[char_int(boggle[i][j])]) {
                str = str + boggle[i][j];
                searchWord(pChild->Child[char_int(boggle[i][j])],
                           boggle, i, j, visited, str);
                str = "";
            }
        }
    }
}
 
// Driver program to test above function
int main()
{
    // Let the given dictionary be following
    char* dictionary[] = { "GEEKS", "FOR", "QUIZ", "GEE" };
 
    // root Node of trie
    TrieNode* root = getNode();
 
    // insert all words of dictionary into trie
    int n = sizeof(dictionary) / sizeof(dictionary[0]);
    for (int i = 0; i < n; i++)
        insert(root, dictionary[i]);
 
    char boggle[M][N] = { { 'G', 'I', 'Z' },
                          { 'U', 'E', 'K' },
                          { 'Q', 'S', 'E' } };
 
    findWords(boggle, root);
 
    return 0;
}

Java




// Java program for Boggle game
public class Boggle {
 
    // Alphabet size
    static final int SIZE = 26;
 
    static final int M = 3;
    static final int N = 3;
 
    // trie Node
    static class TrieNode {
        TrieNode[] Child = new TrieNode[SIZE];
 
        // isLeaf is true if the node represents
        // end of a word
        boolean leaf;
 
        // constructor
        public TrieNode()
        {
            leaf = false;
            for (int i = 0; i < SIZE; i++)
                Child[i] = null;
        }
    }
 
    // If not present, inserts a key into the trie
    // If the key is a prefix of trie node, just
    // marks leaf node
    static void insert(TrieNode root, String Key)
    {
        int n = Key.length();
        TrieNode pChild = root;
 
        for (int i = 0; i < n; i++) {
            int index = Key.charAt(i) - 'A';
 
            if (pChild.Child[index] == null)
                pChild.Child[index] = new TrieNode();
 
            pChild = pChild.Child[index];
        }
 
        // make last node as leaf node
        pChild.leaf = true;
    }
 
    // function to check that current location
    // (i and j) is in matrix range
    static boolean isSafe(int i, int j, boolean visited[][])
    {
        return (i >= 0 && i < M && j >= 0
                && j < N && !visited[i][j]);
    }
 
    // A recursive function to print
    // all words present on boggle
    static void searchWord(TrieNode root, char boggle[][], int i,
                           int j, boolean visited[][], String str)
    {
        // if we found word in trie / dictionary
        if (root.leaf == true)
            System.out.println(str);
 
        // If both I and j in  range and we visited
        // that element of matrix first time
        if (isSafe(i, j, visited)) {
            // make it visited
            visited[i][j] = true;
 
            // traverse all child of current root
            for (int K = 0; K < SIZE; K++) {
                if (root.Child[K] != null) {
                    // current character
                    char ch = (char)(K + 'A');
 
                    // Recursively search reaming character of word
                    // in trie for all 8 adjacent cells of
                    // boggle[i][j]
                    if (isSafe(i + 1, j + 1, visited)
                        && boggle[i + 1][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i, j + 1, visited)
                        && boggle[i][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i, j + 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j + 1, visited)
                        && boggle[i - 1][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i + 1, j, visited)
                        && boggle[i + 1][j] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j,
                                   visited, str + ch);
                    if (isSafe(i + 1, j - 1, visited)
                        && boggle[i + 1][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i, j - 1, visited)
                        && boggle[i][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j - 1, visited)
                        && boggle[i - 1][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j, visited)
                        && boggle[i - 1][j] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j,
                                   visited, str + ch);
                }
            }
 
            // make current element unvisited
            visited[i][j] = false;
        }
    }
 
    // Prints all words present in dictionary.
    static void findWords(char boggle[][], TrieNode root)
    {
        // Mark all characters as not visited
        boolean[][] visited = new boolean[M][N];
        TrieNode pChild = root;
 
        String str = "";
 
        // traverse all matrix elements
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                // we start searching for word in dictionary
                // if we found a character which is child
                // of Trie root
                if (pChild.Child[(boggle[i][j]) - 'A'] != null) {
                    str = str + boggle[i][j];
                    searchWord(pChild.Child[(boggle[i][j]) - 'A'],
                               boggle, i, j, visited, str);
                    str = "";
                }
            }
        }
    }
 
    // Driver program to test above function
    public static void main(String args[])
    {
        // Let the given dictionary be following
        String dictionary[] = { "GEEKS", "FOR", "QUIZ", "GEE" };
 
        // root Node of trie
        TrieNode root = new TrieNode();
 
        // insert all words of dictionary into trie
        int n = dictionary.length;
        for (int i = 0; i < n; i++)
            insert(root, dictionary[i]);
 
        char boggle[][] = { { 'G', 'I', 'Z' },
                            { 'U', 'E', 'K' },
                            { 'Q', 'S', 'E' } };
 
        findWords(boggle, root);
    }
}
// This code is contributed by Sumit Ghosh

Python3




# Python program for Boggle game
class TrieNode:
 
    # Trie node class
    def __init__(self):
        self.children = [None] * 26
 
        # isEndOfWord is True if node represent the end of the word
        self.isEndOfWord = False
 
M = 3
N = 3
class Boogle:
 
    # Trie data structure class
    def __init__(self):
        self.root = self.getNode()
 
    def getNode(self):
 
        # Returns new trie node (initialized to NULLs)
        return TrieNode()
 
    def _charToIndex(self, ch):
 
        # private helper function
        # Converts key current character into index
        # use only 'A' through 'Z' and upper case
        return ord(ch) - ord('A')
 
    def insert(self, key):
 
        # If not present, inserts key into trie
        # If the key is prefix of trie node,
        # just marks leaf node
        pCrawl = self.root
        length = len(key)
        for level in range(length):
            index = self._charToIndex(key[level])
 
            # if current character is not present
            if not pCrawl.children[index]:
 
                pCrawl.children[index] = self.getNode()
            pCrawl = pCrawl.children[index]
            # print('h', self.root.children)
 
        # mark last node as leaf
        pCrawl.isEndOfWord = True
 
 
def is_Safe(i, j, vis):
    return 0 <= i < M and 0 <= j < N and not vis[i][j]
 
def search_word(root, boggle, i, j, vis, string):
    if root.isEndOfWord:
        print(string)
 
    if is_Safe(i, j, vis):
        vis[i][j] = True
        for K in range(26):
            if root.children[K] is not None:
                ch = chr(K+ord('A'))
                 
                # Recursively search reaming character of word
                # in trie for all 8 adjacent cells of boggle[i][j]
                if is_Safe(i + 1, j + 1, vis) and boggle[i + 1][j + 1] == ch:
                    search_word(root.children[K], boggle,
                                i + 1, j + 1, vis, string + ch)
                if is_Safe(i, j + 1, vis) and boggle[i][j + 1] == ch:
                    search_word(root.children[K], boggle,
                                i, j + 1, vis, string + ch)
                if is_Safe(i - 1, j + 1, vis) and boggle[i - 1][j + 1] == ch:
                    search_word(root.children[K], boggle,
                                i - 1, j + 1, vis, string + ch)
                if is_Safe(i + 1, j, vis) and boggle[i + 1][j] == ch:
                    search_word(root.children[K], boggle,
                                i + 1, j, vis, string + ch)
                if is_Safe(i + 1, j - 1, vis) and boggle[i + 1][j - 1] == ch:
                    search_word(root.children[K], boggle,
                                i + 1, j - 1, vis, string + ch)
                if is_Safe(i, j - 1, vis) and boggle[i][j - 1] == ch:
                    search_word(root.children[K], boggle,
                                i, j - 1, vis, string + ch)
                if is_Safe(i - 1, j - 1, vis) and boggle[i - 1][j - 1] == ch:
                    search_word(root.children[K], boggle,
                                i - 1, j - 1, vis, string + ch)
                if is_Safe(i - 1, j, vis) and boggle[i - 1][j] == ch:
                    search_word(root.children[K], boggle,
                                i - 1, j, vis, string + ch)
                vis[i][j] = False
 
 
def char_int(ch):
   
    # private helper function
    # Converts key current character into index
    # use only 'A' through 'Z' and upper case
    return ord(ch) - ord('A')
 
 
def findWords(boggle, root):
   
    # Mark all characters as not visited
    visited = [[False for i in range(N)] for i in range(M)]
 
    pChild = root
 
    string = ""
 
    # traverse all matrix elements
    for i in range(M):
        for j in range(N):
            # we start searching for word in dictionary
            # if we found a character which is child
            # of Trie root
            if pChild.children[char_int(boggle[i][j])]:
                # print('h')
                string = string + boggle[i][j]
                search_word(pChild.children[char_int(boggle[i][j])],
                            boggle, i, j, visited, string)
                string = ""
 
dictionary = ["GEEKS", "FOR", "QUIZ", "GEE"]
 
# root Node of trie
t = Boogle()
 
# insert all words of dictionary into trie
n = len(dictionary)
for i in range(n):
 
    t.insert(dictionary[i])
root = t.root
boggle = [['G', 'I', 'Z'],
          ['U', 'E', 'K'],
          ['Q', 'S', 'E']]
 
# print(root.children)
findWords(boggle, root)
 
# This code is contributed by Yashwant Kumar

C#




// C# program for Boggle game
using System;
 
public class Boggle {
 
    // Alphabet size
    static readonly int SIZE = 26;
 
    static readonly int M = 3;
    static readonly int N = 3;
 
    // trie Node
    public class TrieNode {
        public TrieNode[] Child = new TrieNode[SIZE];
 
        // isLeaf is true if the node represents
        // end of a word
        public bool leaf;
 
        // constructor
        public TrieNode()
        {
            leaf = false;
            for (int i = 0; i < SIZE; i++)
                Child[i] = null;
        }
    }
 
    // If not present, inserts a key into the trie
    // If the key is a prefix of trie node, just
    // marks leaf node
    static void insert(TrieNode root, String Key)
    {
        int n = Key.Length;
        TrieNode pChild = root;
 
        for (int i = 0; i < n; i++) {
            int index = Key[i] - 'A';
 
            if (pChild.Child[index] == null)
                pChild.Child[index] = new TrieNode();
 
            pChild = pChild.Child[index];
        }
 
        // make last node as leaf node
        pChild.leaf = true;
    }
 
    // function to check that current location
    // (i and j) is in matrix range
    static bool isSafe(int i, int j, bool[, ] visited)
    {
        return (i >= 0 && i < M && j >= 0 && j < N && !visited[i, j]);
    }
 
    // A recursive function to print all words present on boggle
    static void searchWord(TrieNode root, char[, ] boggle, int i,
                           int j, bool[, ] visited, String str)
    {
        // if we found word in trie / dictionary
        if (root.leaf == true)
            Console.WriteLine(str);
 
        // If both I and j in range and we visited
        // that element of matrix first time
        if (isSafe(i, j, visited)) {
            // make it visited
            visited[i, j] = true;
 
            // traverse all child of current root
            for (int K = 0; K < SIZE; K++) {
                if (root.Child[K] != null) {
                    // current character
                    char ch = (char)(K + 'A');
 
                    // Recursively search reaming character of word
                    // in trie for all 8 adjacent cells of
                    // boggle[i, j]
                    if (isSafe(i + 1, j + 1, visited) && boggle[i + 1, j + 1] == ch)
                        searchWord(root.Child[K], boggle, i + 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i, j + 1, visited) && boggle[i, j + 1] == ch)
                        searchWord(root.Child[K], boggle, i, j + 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j + 1, visited) && boggle[i - 1, j + 1] == ch)
                        searchWord(root.Child[K], boggle, i - 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i + 1, j, visited) && boggle[i + 1, j] == ch)
                        searchWord(root.Child[K], boggle, i + 1, j,
                                   visited, str + ch);
                    if (isSafe(i + 1, j - 1, visited) && boggle[i + 1, j - 1] == ch)
                        searchWord(root.Child[K], boggle, i + 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i, j - 1, visited) && boggle[i, j - 1] == ch)
                        searchWord(root.Child[K], boggle, i, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j - 1, visited) && boggle[i - 1, j - 1] == ch)
                        searchWord(root.Child[K], boggle, i - 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j, visited) && boggle[i - 1, j] == ch)
                        searchWord(root.Child[K], boggle, i - 1, j,
                                   visited, str + ch);
                }
            }
 
            // make current element unvisited
            visited[i, j] = false;
        }
    }
 
    // Prints all words present in dictionary.
    static void findWords(char[, ] boggle, TrieNode root)
    {
        // Mark all characters as not visited
        bool[, ] visited = new bool[M, N];
        TrieNode pChild = root;
 
        String str = "";
 
        // traverse all matrix elements
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                // we start searching for word in dictionary
                // if we found a character which is child
                // of Trie root
                if (pChild.Child[(boggle[i, j]) - 'A'] != null) {
                    str = str + boggle[i, j];
                    searchWord(pChild.Child[(boggle[i, j]) - 'A'],
                               boggle, i, j, visited, str);
                    str = "";
                }
            }
        }
    }
 
    // Driver program to test above function
    public static void Main(String[] args)
    {
        // Let the given dictionary be following
        String[] dictionary = { "GEEKS", "FOR", "QUIZ", "GEE" };
 
        // root Node of trie
        TrieNode root = new TrieNode();
 
        // insert all words of dictionary into trie
        int n = dictionary.Length;
        for (int i = 0; i < n; i++)
            insert(root, dictionary[i]);
 
        char[, ] boggle = { { 'G', 'I', 'Z' },
                            { 'U', 'E', 'K' },
                            { 'Q', 'S', 'E' } };
        findWords(boggle, root);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
// Javascript program for Boggle game
 
// Alphabet size
let SIZE = 26;
let M = 3;
let N = 3;
 
 // trie Node
class TrieNode
{
    constructor()
    {
        this.leaf=false;
        this.Child = new Array(SIZE);
        for (let i = 0; i < SIZE; i++)
            this.Child[i]=null;
    }
}
 
// If not present, inserts a key into the trie
    // If the key is a prefix of trie node, just
    // marks leaf node
function insert(root,Key)
{
    let n = Key.length;
        let pChild = root;
   
        for (let i = 0; i < n; i++) {
            let index = Key[i].charCodeAt(0) - 'A'.charCodeAt(0);
   
            if (pChild.Child[index] == null)
                pChild.Child[index] = new TrieNode();
   
            pChild = pChild.Child[index];
        }
   
        // make last node as leaf node
        pChild.leaf = true;
}
 
// function to check that current location
    // (i and j) is in matrix range
function isSafe(i,j,visited)
{
    return (i >= 0 && i < M && j >= 0
                && j < N && !visited[i][j]);
}
 
// A recursive function to print
    // all words present on boggle
function searchWord(root,boggle,i,j,visited,str)
{
    // if we found word in trie / dictionary
        if (root.leaf == true)
            document.write(str+" <br>");
   
        // If both I and j in  range and we visited
        // that element of matrix first time
        if (isSafe(i, j, visited)) {
            // make it visited
            visited[i][j] = true;
   
            // traverse all child of current root
            for (let K = 0; K < SIZE; K++) {
                if (root.Child[K] != null) {
                    // current character
                    let ch = String.fromCharCode(K + 'A'.charCodeAt(0));
   
                    // Recursively search reaming character of word
                    // in trie for all 8 adjacent cells of
                    // boggle[i][j]
                    if (isSafe(i + 1, j + 1, visited)
                        && boggle[i + 1][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i, j + 1, visited)
                        && boggle[i][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i, j + 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j + 1, visited)
                        && boggle[i - 1][j + 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j + 1,
                                   visited, str + ch);
                    if (isSafe(i + 1, j, visited)
                        && boggle[i + 1][j] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j,
                                   visited, str + ch);
                    if (isSafe(i + 1, j - 1, visited)
                        && boggle[i + 1][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i + 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i, j - 1, visited)
                        && boggle[i][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j - 1, visited)
                        && boggle[i - 1][j - 1] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j - 1,
                                   visited, str + ch);
                    if (isSafe(i - 1, j, visited)
                        && boggle[i - 1][j] == ch)
                        searchWord(root.Child[K], boggle,
                                   i - 1, j,
                                   visited, str + ch);
                }
            }
   
            // make current element unvisited
            visited[i][j] = false;
        }
}
 
// Prints all words present in dictionary.
function findWords(boggle,root)
{
    // Mark all characters as not visited
        let visited = new Array(M);
        for(let i=0;i<M;i++)
        {
            visited[i]=new Array(N);
            for(let j=0;j<N;j++)
            {
                visited[i][j]=false;
            }
        }
        let pChild = root;
   
        let str = "";
   
        // traverse all matrix elements
        for (let i = 0; i < M; i++) {
            for (let j = 0; j < N; j++) {
                // we start searching for word in dictionary
                // if we found a character which is child
                // of Trie root
                if (pChild.Child[(boggle[i][j]).charCodeAt(0) - 'A'.charCodeAt(0)] != null) {
                    str = str + boggle[i][j];
                    searchWord(pChild.Child[(boggle[i][j]).charCodeAt(0) - 'A'.charCodeAt(0)],
                               boggle, i, j, visited, str);
                    str = "";
                }
            }
        }
}
 
// Driver program to test above function
let dictionary=["GEEKS", "FOR", "QUIZ", "GEE" ];
// root Node of trie
let root = new TrieNode();
 
// insert all words of dictionary into trie
let n = dictionary.length;
for (let i = 0; i < n; i++)
    insert(root, dictionary[i]);
 
let boggle = [[ 'G', 'I', 'Z' ],
[ 'U', 'E', 'K' ],
[ 'Q', 'S', 'E' ]];
 
findWords(boggle, root);
 
 
// This code is contributed by rag2127
</script>

Output: 
 

GEE, GEEKS, QUIZ

Complexity Analysis: 
 

  • Time complexity: O(4^(N^2)). 
    Even after applying trie the time complexity remains same. For every cell there are 4 directions and there are N^2 cells. So the time complexity is O(4^(N^2)).
  • Auxiliary Space: O(N^2). 
    The maximum length of recursion can be N^2, where N is the side of the matrix. So the space Complexity is O(N^2).

This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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