Boggle | Set 2 (Using Trie)

Given a dictionary, a method to do lookup in 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 same cell.


Input: dictionary[] = {"GEEKS", "FOR", "QUIZ", "GO"};
       boggle[][]   = {{'G','I','Z'},
      isWord(str): returns true if str is present in dictionary
                   else false.

Output:  Following words of dictionary are present


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 discus Trie based solution which is better then DFS base solution.
Given Dictionary dictionary[] = {“GEEKS”, “FOR”, “QUIZ”, “GO”}
1. Create an Empty trie and insert all words of given dictionary into trie

      After insertion Tre look like (leaf nodes are in RED )
                    G   F     Q
                 /  |   |     |
                O   E   O     U
                    |   |     |
                    E    R     I
                    |         |  
                    K         Z 

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
   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 seen this element first time then make it visited.
   visited[i][j] = true
      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 c++ implementation of above idea

// C++ program for Boggle game
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 key into trie
// If the key is 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)
                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)
                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];

    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];
                           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'},

    findWords(boggle, root);

    return 0;



This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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