Search a Word in a 2D Grid of characters - GeeksforGeeks


Search a Word in a 2D Grid of characters - GeeksforGeeks
Given a 2D grid of characters and a word, find all occurrences of given word in grid. A word can be matched in all 8 directions at any point. Word is said be found in a direction if all characters match in this direction (not in zig-zag form).
The 8 directions are, Horizontally Left, Horizontally Right, Vertically Up and 4 Diagonal directions.

The idea used here is simple, we check every cell. If cell has first character, then we one by one try all 8 directions from that cell for a match. Implementation is interesting though. We use two arrays x[] and y[] to find next move in all 8 directions.

// Rows and columns in given grid
#define R 3
#define C 14
// For searching in all 8 direction
int x[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int y[] = { -1, 0, 1, -1, 1, -1, 0, 1 };
// This function searches in all 8-direction from point
// (row, col) in grid[][]
bool search2D(char grid[R][C], int row, int col, string word)
{
    // If first character of word doesn't match with
    // given starting point in grid.
    if (grid[row][col] != word[0])
      return false;
    int len = word.length();
    // Search word in all 8 directions starting from (row,col)
    for (int dir = 0; dir < 8; dir++)
    {
        // Initialize starting point for current direction
        int k, rd = row + x[dir], cd = col + y[dir];
        // First character is already checked, match remaining
        // characters
        for (k = 1; k < len; k++)
        {
            // If out of bound break
            if (rd >= R || rd < 0 || cd >= C || cd < 0)
                break;
            // If not matched,  break
            if (grid[rd][cd] != word[k])
                break;
            //  Moving in particular direction
            rd += x[dir], cd += y[dir];
        }
        // If all character matched, then value of must
        // be equal to length of word
        if (k == len)
            return true;
    }
    return false;
}
//  Searches given word in a given matrix in all 8 directions
void patternSearch(char grid[R][C], string word)
{
    // Consider every point as starting point and search
    // given word
    for (int row = 0; row < R; row++)
       for (int col = 0; col < C; col++)
          if (search2D(grid, row, col, word))
             cout << "pattern found at " << row << ", "
                  << col << endl;
}

http://www.geeksforgeeks.org/find-all-occurrences-of-the-word-in-a-matrix/
Given a 2D grid of characters and a word, find all occurrences of given word in grid. A word can be matched in all 8 directions at any point. Word is said be found in a direction if all characters match in this direction (not in zig-zag form).
The 8 directions are, Horizontally Left, Horizontally Right, Vertically Up and 4 Diagonals.
Input:
mat[ROW][COL]= { {'B', 'N', 'E', 'Y', 'S'},
               {'H', 'E', 'D', 'E', 'S'},
          {'S', 'G', 'N', 'D', 'E'}
               };
Word = “DES”

Output:
D(1, 2) E(1, 1) S(2, 0) 
D(1, 2) E(1, 3) S(0, 4) 
D(1, 2) E(1, 3) S(1, 4)
D(2, 3) E(1, 3) S(0, 4)
D(2, 3) E(1, 3) S(1, 4)
D(2, 3) E(2, 4) S(1, 4)
The problem can be easily solved by applying DFS() on each occurrence of first character of the word in the matrix. A cell in 2D matrix can be connected to 8 neighbours. So, unlike standard DFS(), where we recursively call for all adjacent vertices, here we can recursive call for 8 neighbours only.
bool isvalid(int row, int col)
{
    // return true if row number and column number
    // is in range
    return (row >= 0) && (row < ROW) &&
           (col >= 0) && (col < COL);
}
 
// These arrays are used to get row and column
// numbers of 8 neighboursof a given cell
int rowNum[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int colNum[] = {-1, 0, 1, -1, 1, -1, 0, 1};
 
// A utility function to do DFS for a 2D boolean
// matrix. It only considers the 8 neighbours as
// adjacent vertices
void DFS(char mat[][COL], int row, int col, char* word,
        string path, int index, int n)
{
    // return if current character doesn't match with
    // the next character in the word
    if (index > n || mat[row][col] != word[index])
        return;
 
    //append current character position to path
    path += string(1, word[index]) + "(" + to_string(row)
             + ", " + to_string(col) + ") ";
 
    // current character matches with the last character
    // in the word
    if (index == n)
    {
        cout << path << endl;
        return;
    }
 
    // Recur for all connected neighbours
    for (int k = 0; k < 8; ++k)
        if (isvalid(row + rowNum[k], col + colNum[k]))
            DFS(mat, row + rowNum[k], col + colNum[k],
               word, path, index+1, n);
}
 
// The main function to find all occurrences of the
// word in a matrix
void findWords(char mat[][COL], char* word, int n)
{
    // traverse through the all cells of given matrix
    for (int i = 0; i < ROW; ++i)
        for (int j = 0; j < COL; ++j)
 
            // occurrence of first character in matrix
            if (mat[i][j] == word[0])
 
                // check and print if path exists
                DFS(mat, i, j, word, "", 0, n);
}
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