# Boggle | Set 2 (Using Trie)

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: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 then 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, ‘Q’ boggle (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 ` `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); ` `    ``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 `

## 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 `

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.