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.
Example:
Input: dictionary[] = {"GEEKS", "FOR", "QUIZ", "GO"};
boggle[][] = {{'G','I','Z'},
{'U','E','K'},
{'Q','S','E'}};
isWord(str): returns true if str is present in dictionary
else false.
Output: Following words of dictionary are present
GEEKS
QUIZ
The idea is to consider every character as a starting character and find all words starting with it. All words starting from a character can be found using Depth First Traversal. We do depth first traversal starting from every cell. We keep track of visited cells to make sure that a cell is considered only once in a word.
// C++ program for Boggle game #include<iostream> #include<cstring> using namespace std; #define M 3 #define N 3 // Let the given dictionary be following string dictionary[] = {"GEEKS", "FOR", "QUIZ", "GO"}; int n = sizeof(dictionary)/sizeof(dictionary[0]); // A given function to check if a given string is present in // dictionary. The implementation is naive for simplicity. As // per the question dictionary is given to us. bool isWord(string &str) { // Linearly search all words for (int i=0; i<n; i++) if (str.compare(dictionary[i]) == 0) return true; return false; } // A recursive function to print all words present on boggle void findWordsUtil(char boggle[M][N], bool visited[M][N], int i, int j, string &str) { // Mark current cell as visited and append current character // to str visited[i][j] = true; str = str + boggle[i][j]; // If str is present in dictionary, then print it if (isWord(str)) cout << str << endl; // Traverse 8 adjacent cells of boggle[i][j] for (int row=i-1; row<=i+1 && row<M; row++) for (int col=j-1; col<=j+1 && col<N; col++) if (row>=0 && col>=0 && !visited[row][col]) findWordsUtil(boggle,visited, row, col, str); // Erase current character from string and mark visited // of current cell as false str.erase(str.length()-1); visited[i][j] = false; } // Prints all words present in dictionary. void findWords(char boggle[M][N]) { // Mark all characters as not visited bool visited[M][N] = {{false}}; // Initialize current string string str = ""; // Consider every character and look for all words // starting with this character for (int i=0; i<M; i++) for (int j=0; j<N; j++) findWordsUtil(boggle, visited, i, j, str); } // Driver program to test above function int main() { char boggle[M][N] = {{'G','I','Z'}, {'U','E','K'}, {'Q','S','E'}}; cout << "Following words of dictionary are present
"; findWords(boggle); return 0; } |
Output:
Following words of dictionary are present GEEKS QUIZ
Note that the above solution may print the same word multiple times. For example, if we add “SEEK” to the dictionary, it is printed multiple times. To avoid this, we can use hashing to keep track of all printed words.
In below set 2, we have discussed Trie based optimized solution:
Boggle | Set 2 (Using Trie)
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