Search a Word in a 2D Grid of characters

3.1

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.

Example:

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

Output: pattern found at 0, 0
        pattern found at 0, 8
        pattern found at 1, 0

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

Below diagram shows a bigger grid and presence of different words in it.
wordsearch(1)

Source: Microsoft Interview Question

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 is C++ implementation of the same.

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

// 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;
}

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

    patternSearch(grid, "GEEKS");
    cout << endl; 
    patternSearch(grid, "EEE");
    return 0;
}

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

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.

Asked in: Microsoft, Zoho

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

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