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Check if a word exists in a grid or not

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  • Difficulty Level : Medium
  • Last Updated : 23 Jan, 2023
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Given a 2D grid of characters and a word/ multiple words, the task is to check if that word/words exist in the grid or not. A word can be matched in 4 directions at any point.
The 4 directions are Horizontally Left and Right, Vertically Up and Down. 
Examples: 

Input:  grid[][] = {"axmy",
                    "bgdf",
                    "xeet",
                    "raks"};
Output: Yes

a x m y
b g d f
x e e t
r a k s

Input: grid[][] = {"axmy",
                   "brdf",
                   "xeet",
                   "rass"};
Output : No

Source: Microsoft Interview

Approach when a single word is to be checked : The idea used here is described in the steps below: 

  • Check every cell, if the cell has the first character, then recur one by one and try all 4 directions from that cell for a match.
  • Mark the position in the grid as visited and recur in the 4 possible directions.
  • After recurring, again mark the position as unvisited.
  • Once all the letters in the word are matched, return true.

Below is the implementation of the above approach:  

C++




// C++ program to check if the word
// exists in the grid or not
#include <bits/stdc++.h>
using namespace std;
#define r 4
#define c 5
 
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
bool findmatch(char mat[r], string pat, int x, int y,
               int nrow, int ncol, int level)
{
    int l = pat.length();
 
    // Pattern matched
    if (level == l)
        return true;
 
    // Out of Boundary
    if (x < 0 || y < 0 || x >= nrow || y >= ncol)
        return false;
 
    // If grid matches with a letter while
    // recursion
    if (mat[x][y] == pat[level]) {
 
        // Marking this cell as visited
        char temp = mat[x][y];
        mat[x][y] = '#';
 
        // finding subpattern in 4 directions
        bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
                   findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
                   findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
                   findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
 
        // marking this cell
        // as unvisited again
        mat[x][y] = temp;
        return res;
    }
    else // Not matching then false
        return false;
}
 
// Function to check if the word exists in the grid or not
bool checkMatch(char mat[r], string pat, int nrow, int ncol)
{
 
    int l = pat.length();
 
    // if total characters in matrix is
    // less than pattern length
    if (l > nrow * ncol)
        return false;
 
    // Traverse in the grid
    for (int i = 0; i < nrow; i++) {
        for (int j = 0; j < ncol; j++) {
 
            // If first letter matches, then recur and check
            if (mat[i][j] == pat[0])
                if (findmatch(mat, pat, i, j, nrow, ncol, 0))
                    return true;
        }
    }
    return false;
}
 
// Driver Code
int main()
{
    char grid[r] = { "axmy",
                        "bgdf",
                        "xeet",
                        "raks" };
 
    // Function to check if word exists or not
    if (checkMatch(grid, "geeks", r, c))
        cout << "Yes";
    else
        cout << "No";
 
 return 0;
 
}

Java




// Java program to check if the word
// exists in the grid or not
class GFG
{
     
static final int r = 4;
static final int c = 4;
 
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
static boolean findmatch(char mat[][], String pat, int x, int y,
                        int nrow, int ncol, int level)
{
    int l = pat.length();
 
    // Pattern matched
    if (level == l)
        return true;
 
    // Out of Boundary
    if (x < 0 || y < 0 || x >= nrow || y >= ncol)
        return false;
 
    // If grid matches with a letter while
    // recursion
    if (mat[x][y] == pat.charAt(level))
    {
 
        // Marking this cell as visited
        char temp = mat[x][y];
        mat[x][y] = '#';
 
        // finding subpattern in 4 directions
        boolean res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
 
        // marking this cell
        // as unvisited again
        mat[x][y] = temp;
        return res;
    }
    else // Not matching then false
        return false;
}
 
// Function to check if the word exists in the grid or not
static boolean checkMatch(char mat[][], String pat, int nrow, int ncol)
{
 
    int l = pat.length();
 
    // if total characters in matrix is
    // less than pattern length
    if (l > nrow * ncol)
        return false;
 
    // Traverse in the grid
    for (int i = 0; i < nrow; i++)
    {
        for (int j = 0; j < ncol; j++)
        {
 
            // If first letter matches, then recur and check
            if (mat[i][j] == pat.charAt(0))
                if (findmatch(mat, pat, i, j, nrow, ncol, 0))
                    return true;
        }
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    char grid[][] = { "axmy".toCharArray(),
                        "bgdf".toCharArray(),
                        "xeet".toCharArray(),
                        "raks".toCharArray() };
 
    // Function to check if word exists or not
    if (checkMatch(grid, "geeks", r, c))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to check if the word
# exists in the grid or not
 
r = 4
c = 4
 
# Function to check if a word exists
# in a grid starting from the first
# match in the grid level: index till 
# which pattern is matched x, y: current
# position in 2D array
def findmatch(mat, pat, x, y,
              nrow, ncol, level) :
 
    l = len(pat)
 
    # Pattern matched
    if (level == l) :
        return True
 
    # Out of Boundary
    if (x < 0 or y < 0 or
        x >= nrow or y >= ncol) :
        return False
 
    # If grid matches with a letter
    # while recursion
    if (mat[x][y] == pat[level]) :
 
        # Marking this cell as visited
        temp = mat[x][y]
        mat[x].replace(mat[x][y], "#")
 
        # finding subpattern in 4 directions
        res = (findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
               findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
               findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
               findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1))
 
        # marking this cell as unvisited again
        mat[x].replace(mat[x][y], temp)
        return res
     
    else : # Not matching then false
        return False
 
# Function to check if the word
# exists in the grid or not
def checkMatch(mat, pat, nrow, ncol) :
 
    l = len(pat)
 
    # if total characters in matrix is
    # less than pattern length
    if (l > nrow * ncol) :
        return False
 
    # Traverse in the grid
    for i in range(nrow) :
        for j in range(ncol) :
 
            # If first letter matches, then
            # recur and check
            if (mat[i][j] == pat[0]) :
                if (findmatch(mat, pat, i, j,
                              nrow, ncol, 0)) :
                    return True
    return False
 
# Driver Code
if __name__ == "__main__" :
 
    grid = ["axmy", "bgdf",
            "xeet", "raks"]
 
    # Function to check if word
    # exists or not
    if (checkMatch(grid, "geeks", r, c)) :
        print("Yes")
    else :
        print("No")
 
# This code is contributed by Ryuga

C#




// C# program to check if the word
// exists in the grid or not
using System;
 
class GFG
{
     
static readonly int r = 4;
static readonly int c = 4;
 
// Function to check if a word exists in a grid
// starting from the first match in the grid
// level: index till which pattern is matched
// x, y: current position in 2D array
static bool findmatch(char [,]mat, String pat, int x, int y,
                        int nrow, int ncol, int level)
{
    int l = pat.Length;
 
    // Pattern matched
    if (level == l)
        return true;
 
    // Out of Boundary
    if (x < 0 || y < 0 || x >= nrow || y >= ncol)
        return false;
 
    // If grid matches with a letter while
    // recursion
    if (mat[x, y] == pat[level])
    {
 
        // Marking this cell as visited
        char temp = mat[x, y];
        mat[x, y] = '#';
 
        // finding subpattern in 4 directions
        bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
                    findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
 
        // marking this cell
        // as unvisited again
        mat[x, y] = temp;
        return res;
    }
    else // Not matching then false
        return false;
}
 
// Function to check if the word exists in the grid or not
static bool checkMatch(char [,]mat, String pat, int nrow, int ncol)
{
 
    int l = pat.Length;
 
    // if total characters in matrix is
    // less than pattern length
    if (l > nrow * ncol)
        return false;
 
    // Traverse in the grid
    for (int i = 0; i < nrow; i++)
    {
        for (int j = 0; j < ncol; j++)
        {
 
            // If first letter matches, then recur and check
            if (mat[i, j] == pat[0])
                if (findmatch(mat, pat, i, j, nrow, ncol, 0))
                    return true;
        }
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    char [,]grid = { {'a','x','m','y'},
                    {'b','g','d','f'},
                    {'x','e','e','t'},
                    {'r','a','k','s'} };
 
    // Function to check if word exists or not
    if (checkMatch(grid, "geeks", r, c))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // JavaScript program to check if the word
    // exists in the grid or not
     
    let r = 4;
    let c = 4;
 
    // Function to check if a word exists in a grid
    // starting from the first match in the grid
    // level: index till which pattern is matched
    // x, y: current position in 2D array
    function findmatch(mat, pat, x, y, nrow, ncol, level)
    {
        let l = pat.length;
 
        // Pattern matched
        if (level == l)
            return true;
 
        // Out of Boundary
        if (x < 0 || y < 0 || x >= nrow || y >= ncol)
            return false;
 
        // If grid matches with a letter while
        // recursion
        if (mat[x][y] == pat[level])
        {
 
            // Marking this cell as visited
            let temp = mat[x][y];
            mat[x][y] = '#';
 
            // finding subpattern in 4 directions
            let res =
            findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) |
            findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) |
            findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) |
            findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1);
 
            // marking this cell
            // as unvisited again
            mat[x][y] = temp;
            return res;
        }
        else // Not matching then false
            return false;
    }
 
    // Function to check if the word exists in the grid or not
    function checkMatch(mat, pat, nrow, ncol)
    {
 
        let l = pat.length;
 
        // if total characters in matrix is
        // less than pattern length
        if (l > nrow * ncol)
            return false;
 
        // Traverse in the grid
        for (let i = 0; i < nrow; i++)
        {
            for (let j = 0; j < ncol; j++)
            {
 
                // If first letter matches, then recur and check
                if (mat[i][j] == pat[0])
                    if (findmatch(mat, pat, i, j, nrow, ncol, 0))
                        return true;
            }
        }
        return false;
    }
     
    let grid = [ "axmy".split(''),
                        "bgdf".split(''),
                        "xeet".split(''),
                        "raks".split('') ];
   
    // Function to check if word exists or not
    if (checkMatch(grid, "geeks", r, c))
        document.write("Yes");
    else
        document.write("No");
 
</script>

Output

Yes

Time Complexity: O(r*c), as we are using recursion to traverse the matrix. Where r and c are the rows and columns of the grid.

Auxiliary Space: O(r*c), as we are using extra space for the matrix. Where r and c are the rows and columns of the grid.

Approach when a group of words are to be checked : The idea used here is described in the steps below: 

  • iterate through group of words and check every cell, if the cell has the first character, then recur one by one and try all 4 directions from that cell for a match.
  • Mark the position in the grid as visited and recur in the 4 possible directions.
  • After recurring, again mark the position as unvisited.
  • Once all the letters in the word are matched, return true, put it in answer list.
  • return the answer list from a function and display 

C++




#include <bits/stdc++.h>
using namespace std;
// making a solution class to solve the problem and to keep
// the components and functions of solution together
class Solution {
public:
    // making the possible moves in movers array
    // if 4 directions are to be checked
    vector<vector<int> > mover
        = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };
    // if 8 directions are to be checked
    // vector<vector<int>>mover={{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
    // making the board global variable
    vector<vector<char> > board;
    // depth first search for the string, with the
    // coordinates and a visited array to take care that we
    // do not overlap the places visited already
    bool dfs(int x, int y, string& s,
             vector<vector<bool> > vis)
    {
        // if string length becomes 0 means string is found
        if (s.length() == 0)
            return true;
        vis[x][y] = true;
        // making a solution boolean to see if we can
        // perform depth search to find answer
        bool sol = false;
        // making possible moves
        for (int i = 0; i < mover.size(); i++) {
            int curr_x = mover[i][0] + x;
            int curr_y = mover[i][1] + y;
            // checking for out of bound areas
            if (curr_x >= 0 && curr_x < board.size()) {
                if (curr_y >= 0
                    && curr_y < board[0].size()) {
                    // checking for similarity in the first
                    // letter and the visited array
                    if (board[curr_x][curr_y] == s[0]
                        && vis[curr_x][curr_y] == false) {
                        string k = s.substr(
                            1); // removing the first letter
                                // from the string
                        sol |= dfs(curr_x, curr_y, k, vis);
                    }
                }
            }
        }
        return sol;
    }
    // making a function findwords to find words along with
    // their location which inputs the board and list of
    // words
    vector<string> findWords(vector<vector<char> >& board,
                             vector<string>& words)
    {
        this->board
            = board; // making board a global variable
        vector<string> ans;
        vector<vector<bool> > vis(
            board.size(),
            vector<bool>(board[0].size(),
                         false)); // visited array
        for (auto& word : words) {
            for (int i = 0; i < board.size(); i++) {
                for (int j = 0; j < board[i].size(); j++) {
                    if (board[i][j] == word[0]) {
                        // if first letter of(i,j)==
                        // string's first letter then we can
                        // perform dfs to check the
                        // possiblity of string being present
                        // from location (i,j)
                        string s = word.substr(1);
                        if (dfs(i, j, s, vis)) {
                            ans.push_back(word);
                            break;
                        }
                    }
                }
                if (ans.size() && ans.back() == word)
                    break;
            }
        }
        return ans;
    }
};
int main()
{
    // making 1 instance of class solution as solver
    Solution solver;
    vector<vector<char> > board
        = { { 'o', 'a', 'a', 'n' },
            { 'e', 't', 'a', 'e' },
            { 'i', 'h', 'k', 'r' },
            { 'i', 'f', 'l', 'v' } };
    vector<string> words = { "oath", "pea", "eat", "rain" };
    // using the function findwords from our solution class
    // to find the answer
    vector<string> ans = solver.findWords(board, words);
    // printing the answer
    cout << "words present:\n";
    for (auto& part : ans)
        cout << part << endl;
    return 0;
}//contributed by Naman Anand

Python3




class Solution:
    # making the possible moves in movers array
    # if 4 directions are to be checked
    mover = [ [1, 0], [0, 1], [-1, 0], [0, -1] ]
    # if 8 directions are to be checked
    # mover = [[1,0], [0,1], [-1,0], [0,-1], [1,1], [-1,-1], [1,-1], [-1,1]]
    # making the board global variable
    board = []
    # depth first search for the string, with the
    # coordinates and a visited array to take care that we
    # do not overlap the places visited already
    def dfs(self, x, y, s, vis):
        # if string length becomes 0 means string is found
        if len(s) == 0:
            return True
        vis[x][y] = True
        # making a solution boolean to see if we can
        # perform depth search to find answer
        sol = False
        # making possible moves
        for i in range(len(self.mover)):
            curr_x = self.mover[i][0] + x
            curr_y = self.mover[i][1] + y
            # checking for out of bound areas
            if curr_x >= 0 and curr_x < len(self.board):
                if curr_y >= 0 and curr_y < len(self.board[0]):
                    # checking for similarity in the first
                    # letter and the visited array
                    if self.board[curr_x][curr_y] == s[0] and vis[curr_x][curr_y] == False:
                        # removing the first letter from the string
                        k = s[1:]
                        sol |= self.dfs(curr_x, curr_y, k, vis)
        return sol
 
    # making a function findwords to find words along with
    # their location which inputs the board and list of
    # words
    def findWords(self, board, words):
        self.board = board
        ans = []
        for word in words:
            vis = [[False for _ in range(len(board[0]))] for _ in range(len(board))]
            for i in range(len(board)):
                for j in range(len(board[i])):
                    if board[i][j] == word[0]:
                        # if first letter of(i,j)==
                        # string's first letter then we can
                        # perform dfs to check the
                        # possiblity of string being present
                        # from location (i,j)
                        s = word[1:]
                        if self.dfs(i, j, s, vis):
                            ans.append(word)
                            break
                if ans and ans[-1] == word:
                    break
        return ans
 
# making 1 instance of class solution as solver
solver = Solution()
board = [ ['o','a','a','n'], ['e','t','a','e'], ['i','h','k','r'], ['i','f','l','v'] ]
words = ["oath","pea","eat","rain"]
 
#Function call
print("Words present: ")
arr=solver.findWords(board, words)
for i in range(len(arr)):
  print(arr[i])
 
  # This code is contributed by lokeshpotta20.

Output

words present:
oath
eat

Time Complexity: O(r*c*len(words)*number of words), as we are using recursion to traverse the matrix. Where r and c are the rows and columns of the grid.

Auxiliary Space: O(r*c*number of words), as we are using extra space for the matrix. Where r and c are the rows and columns of the grid.


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