Search a Word in a 2D Grid of characters

15 Feb, 2022
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++
Java
Python3
C#
Javascript
// C++ programs to search a word in a 2D grid#include <bits/stdc++.h>using namespace std; // For searching in all 8 directionint 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 directionsvoid 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 programint 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 program to search// a word in a 2D gridimport 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 program to search a word in a 2D gridclass 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 Codeif __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# program to search a word in a 2D gridusing 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
<script> // JavaScript program to search// a word in a 2D grid  // Rows and columns in the given gridlet R, C; // For searching in all 8 directionlet 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 directionsfunction 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 codeR = 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. 
rachana soma
Rajput-Ji
yezhengli9
rbaikejiang
andrew1234
coderangshu
shivam2608
avanitrachhadiya2155
harshitkap00r
sudhanshublaze
Amazon
FactSet
Microsoft
Morgan Stanley
Samsung
Zoho
Matrix
Pattern Searching
Zoho
Morgan Stanley
Amazon
Microsoft
Samsung
FactSet
Matrix
Pattern Searching
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