Search a Word in a 2D Grid of characters

Given a 2D grid of characters and a word, find all occurrences of the given word in the grid. A word can be matched in all 8 directions at any point. Word is said to 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, Vertically Down and 4 Diagonal directions.
Example:

Input:  grid[][] = {"GEEKSFORGEEKS",
                    "GEEKSQUIZGEEK",
                    "IDEQAPRACTICE"};
        word = "GEEKS"

Output: pattern found at 0, 0
        pattern found at 0, 8
        pattern found at 1, 0
Explanation: 'GEEKS' can be found as prefix of
1st 2 rows and suffix of first row

Input:  grid[][] = {"GEEKSFORGEEKS",
                    "GEEKSQUIZGEEK",
                    "IDEQAPRACTICE"};
        word = "EEE"

Output: pattern found at 0, 2
        pattern found at 0, 10
        pattern found at 2, 2
        pattern found at 2, 12
Explanation: EEE can be found in first row 
twice at index 2 and index 10
and in second row at 2 and 12

Below diagram shows a bigger grid and presence of different words in it. 

Source: Microsoft Interview Question.

Approach: 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. 
Below are implementation of the same:  

C++

// C++ programs to search a word in a 2D grid
#include <bits/stdc++.h>
using namespace std;

// 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, int row, int col,
               string word, int R, int C)
{
    // If first character of word doesn't
    // match with given starting point in grid.
    if (*(grid+row*C+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*C+cd) != word[k])
                break;

            // Moving in particular direction
            rd += x[dir], cd += y[dir];
        }

        // If all character matched, then value of k 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, string word,
                  int R, int C)
{
    // 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, R, C))
                cout << "pattern found at "
                     << row << ", "
                     << col << endl;
}

// Driver program
int main()
{
      int R = 3, C = 13;
    char grid[R][C] = { "GEEKSFORGEEKS",
                        "GEEKSQUIZGEEK",
                        "IDEQAPRACTICE" };

    patternSearch((char *)grid, "GEEKS", R, C);
    cout << endl;
    patternSearch((char *)grid, "EEE", R, C);
    return 0;
}

Java

// Java program to search
// a word in a 2D grid
import java.io.*;
import java.util.*;

class GFG {

    // Rows and columns in the given grid
    static int R, C;

    // For searching in all 8 direction
    static int[] x = { -1, -1, -1, 0, 0, 1, 1, 1 };
    static int[] y = { -1, 0, 1, -1, 1, -1, 0, 1 };

    // This function searches in all
    // 8-direction from point
    // (row, col) in grid[][]
    static boolean search2D(char[][] grid, 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.charAt(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.charAt(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
    static void patternSearch(
        char[][] grid,
        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 (grid[row][col]==word.charAt(0)  &&
                    search2D(grid, row, col, word))
                        System.out.println(
                            "pattern found at " + row + ", " + col);
            }
        }
    }

    // Driver code
    public static void main(String args[])
    {
        R = 3;
        C = 13;
        char[][] grid = { { 'G', 'E', 'E', 'K', 'S', 'F', 'O', 'R', 'G', 'E', 'E', 'K', 'S' },
                          { 'G', 'E', 'E', 'K', 'S', 'Q', 'U', 'I', 'Z', 'G', 'E', 'E', 'K' },
                          { 'I', 'D', 'E', 'Q', 'A', 'P', 'R', 'A', 'C', 'T', 'I', 'C', 'E' } };
        patternSearch(grid, "GEEKS");
        System.out.println();
        patternSearch(grid, "EEE");
    }
}

// This code is contributed by rachana soma

Python3

# Python3 program to search a word in a 2D grid
class GFG:

    def __init__(self):
        self.R = None
        self.C = None
        self.dir = [[-1, 0], [1, 0], [1, 1],
                    [1, -1], [-1, -1], [-1, 1],
                    [0, 1], [0, -1]]

    # This function searches in all 8-direction
    # from point(row, col) in grid[][]
    def search2D(self, grid, row, col, word):

        # If first character of word doesn't match
        # with the given starting point in grid.
        if grid[row][col] != word[0]:
            return False

        # Search word in all 8 directions
        # starting from (row, col)
        for x, y in self.dir:

            # Initialize starting point
            # for current direction
            rd, cd = row + x, col + y
            flag = True

            # First character is already checked,
            # match remaining characters
            for k in range(1, len(word)):

                # If out of bound or not matched, break
                if (0 <= rd <self.R and
                    0 <= cd < self.C and
                    word[k] == grid[rd][cd]):

                    # Moving in particular direction
                    rd += x
                    cd += y
                else:
                    flag = False
                    break

            # If all character matched, then
            # value of flag must be false
            if flag:
                return True
        return False

    # Searches given word in a given matrix
    # in all 8 directions
    def patternSearch(self, grid, word):

        # Rows and columns in given grid
        self.R = len(grid)
        self.C = len(grid[0])

        # Consider every point as starting point
        # and search given word
        for row in range(self.R):
            for col in range(self.C):
                if self.search2D(grid, row, col, word):
                    print("pattern found at " +
                           str(row) + ', ' + str(col))

# Driver Code
if __name__=='__main__':
    grid = ["GEEKSFORGEEKS",
            "GEEKSQUIZGEEK",
            "IDEQAPRACTICE"]
    gfg = GFG()
    gfg.patternSearch(grid, 'GEEKS')
    print('')
    gfg.patternSearch(grid, 'EEE')

# This code is contributed by Yezheng Li

C#

// C# program to search a word in a 2D grid
using System;
class GFG {

    // Rows and columns in given grid
    static int R, C;

    // For searching in all 8 direction
    static int[] x = { -1, -1, -1, 0, 0, 1, 1, 1 };
    static int[] y = { -1, 0, 1, -1, 1, -1, 0, 1 };

    // This function searches in all 8-direction
    // from point (row, col) in grid[, ]
    static bool search2D(char[, ] grid, 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 k
            // 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
    static void patternSearch(char[, ] grid,
                              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)) {
                    Console.WriteLine("pattern found at " + row + ", " + col);
                }
            }
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        R = 3;
        C = 13;
        char[, ] grid = { { 'G', 'E', 'E', 'K', 'S', 'F', 'O',
                            'R', 'G', 'E', 'E', 'K', 'S' },
                          { 'G', 'E', 'E', 'K', 'S', 'Q', 'U',
                            'I', 'Z', 'G', 'E', 'E', 'K' },
                          { 'I', 'D', 'E', 'Q', 'A', 'P', 'R',
                            'A', 'C', 'T', 'I', 'C', 'E' } };
        patternSearch(grid, "GEEKS");
        Console.WriteLine();
        patternSearch(grid, "EEE");
    }
}

#This code is contributed by Rajput - Ji

Javascript

<script>

// JavaScript program to search
// a word in a 2D grid

 // Rows and columns in the given grid
let R, C;

// For searching in all 8 direction
let x=[-1, -1, -1, 0, 0, 1, 1, 1];

let y=[-1, 0, 1, -1, 1, -1, 0, 1];

// This function searches in all
    // 8-direction from point
    // (row, col) in grid[][]
function search2D(grid,row,col,word)
{
    // If first character of word
        // doesn't match with
        // given starting point in grid.
        if (grid[row][col] != word[0])
            return false;

        let len = word.length;

        // Search word in all 8 directions
        // starting from (row, col)
        for (let dir = 0; dir < 8; dir++) {
            // Initialize starting point
            // for current direction
            let 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
function patternSearch( grid,word)
{
    // Consider every point as starting
        // point and search given word
        for (let row = 0; row < R; row++) {
            for (let col = 0; col < C; col++) {
                if (search2D(grid, row, col, word))
                    document.write(
                        "pattern found at " + row + ", " + col+"<br>");
            }
        }
}

// Driver code
R = 3;
C = 13;
let grid = [[ 'G', 'E', 'E', 'K', 'S', 'F', 'O',
'R', 'G', 'E', 'E', 'K', 'S' ],

[ 'G', 'E', 'E', 'K', 'S', 'Q', 'U', 'I', 'Z',
'G', 'E', 'E', 'K' ],

[ 'I', 'D', 'E', 'Q', 'A', 'P', 'R', 'A', 'C',
'T', 'I', 'C', 'E' ] ];
patternSearch(grid, "GEEKS");
document.write("<br>");
patternSearch(grid, "EEE");


// This code is contributed by avanitrachhadiya2155
</script>

Output:

pattern found at 0, 0
pattern found at 0, 8
pattern found at 1, 0

pattern found at 0, 2
pattern found at 0, 10
pattern found at 2, 2
pattern found at 2, 12

Complexity Analysis:

• Time complexity: O(R*C*8*len(str)). 
  All the cells will be visited and traversed in all 8 directions, where R and C is side of matrix so time complexity is O(R*C).
• Auxiliary Space: O(1). 
  As no extra space is needed.

Exercise: The above solution only print locations of word. Extend it to print the direction where word is present.
See this for solution of exercise.
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.