Skip to content
Related Articles

Related Articles

Improve Article

Check if a given string can be formed using characters of adjacent cells of a Matrix

  • Difficulty Level : Basic
  • Last Updated : 03 Jun, 2021

Given a matrix board of characters and a string Word, the task is to check if Word exists on the board constructed from a sequence of horizontally and vertically adjacent characters only. Each character can be used only once.

Examples: 

Input: 
board = { {‘A’, ‘B’, ‘C’, ‘E’}, {‘S’, ‘F’, ‘C’, ‘S’}, {‘A’, ‘D’, ‘E’, ‘E’} } 
Word = “SEE” 
Output: True 
Explanation: “SEE” can be formed using characters at (1, 3)[S], (2, 3)[E] and (2, 2)[E].

Input: 
board = { {‘A’, ‘B’, ‘C’, ‘E’}, {‘S’, ‘F’, ‘C’, ‘S’}, {‘A’, ‘D’, ‘E’, ‘E’} } 
Word = “ABCB” 
Output: False 
Explanation: “ABCB” can not be formed by using adjacent characters without repetition. 
 

Approach: The approach to solving this problem is to traverse all the characters in the matrix and find the occurrence of the first character of the word. Whenever found, recursively keep checking its adjacent horizontal and vertical cells for the next character. Repeat this process until all the characters are found one by one. Any instance where all the characters match signifies Word is found. If no such instance occurs, Word is not found. 



Below is the implementation of the above logic: 

C++




// C++ Program to check if a given
// word can be formed from the
// adjacent characters in a matrix
// of characters
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the word exists
bool checkWord(vector<vector<char> >& board,
               string& word, int index,
               int row, int col)
{
    // If index exceeds board range
    if (row < 0 || col < 0
        || row >= board.size()
        || col >= board[0].size())
        return false;
 
    // If the current cell does not
    // contain the required character
    if (board[row][col] != word[index])
        return false;
 
    // If the cell contains the required
    // character and is the last character
    // of the word required to be matched
    else if (index == word.size() - 1)
 
        // Return true as word is found
        return true;
 
    char temp = board[row][col];
 
    // Mark cell visited
    board[row][col] = '*';
 
    // Check Adjacent cells
    // for the next character
    if (checkWord(board, word,
                  index + 1, row + 1, col)
        || checkWord(board, word,
                     index + 1, row - 1, col)
        || checkWord(board, word,
                     index + 1, row, col + 1)
        || checkWord(board, word,
                     index + 1, row, col - 1)) {
 
        board[row][col] = temp;
 
        return true;
    }
 
    // Restore cell value
    board[row][col] = temp;
    return false;
}
 
// Driver Code
int main()
{
    vector<vector<char> > board
        = { { 'A', 'B', 'C', 'E' },
            { 'S', 'F', 'C', 'S' },
            { 'A', 'D', 'E', 'E' } };
    string word = "CFDASABCESEE";
 
    for (int i = 0; i < board.size(); i++) {
        for (int j = 0; j < board[0].size(); j++) {
 
            if (board[i][j] == word[0]
                && checkWord(
                       board, word,
                       0, i, j)) {
 
                cout << "True" << '\n';
                return 0;
            }
        }
    }
    cout << "False" << '\n';
    return 0;
}

Java




// Java Program to check if a given
// word can be formed from the
// adjacent characters in a matrix
// of characters
import java.util.*;
class GFG{
 
// Function to check if the word exists
static boolean checkWord(char [][]board,
                 String word, int index,
                       int row, int col)
{
    // If index exceeds board range
    if (row < 0 || col < 0 ||
        row >= board.length ||
        col >= board[0].length)
        return false;
 
    // If the current cell does not
    // contain the required character
    if (board[row][col] != word.charAt(index))
        return false;
 
    // If the cell contains the required
    // character and is the last character
    // of the word required to be matched
    else if (index == word.length() - 1)
 
        // Return true as word is found
        return true;
 
    char temp = board[row][col];
 
    // Mark cell visited
    board[row][col] = '*';
 
    // Check Adjacent cells
    // for the next character
    if (checkWord(board, word,
                  index + 1, row + 1, col) ||
        checkWord(board, word,
                  index + 1, row - 1, col) ||
        checkWord(board, word,
                  index + 1, row, col + 1) ||
        checkWord(board, word,
                  index + 1, row, col - 1))
    {
        board[row][col] = temp;
 
        return true;
    }
 
    // Restore cell value
    board[row][col] = temp;
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    char[][] board = { { 'A', 'B', 'C', 'E' },
                       { 'S', 'F', 'C', 'S' },
                       { 'A', 'D', 'E', 'E' } };
    String word = "CFDASABCESEE";
 
    for (int i = 0; i < board.length; i++)
    {
        for (int j = 0; j < board[0].length; j++)
        {
            if (board[i][j] == word.charAt(0) &&
                checkWord(board, word, 0, i, j))
            {
                System.out.println("True");
                return;
            }
        }
    }
    System.out.println("False");
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python 3 Program to check if a given
# word can be formed from the
# adjacent characters in a matrix
# of characters
 
# Function to check if
# the word exists
def checkWord(board, word,
              index, row, col):
 
    # If index exceeds board range
    if (row < 0 or col < 0 or
        row >= len(board) or
        col >= len(board[0])):
        return False
 
    # If the current cell does not
    # contain the required character
    if (board[row][col] != word[index]):
        return False
 
    # If the cell contains the required
    #character and is the last character
    # of the word required to be matched
    elif (index == len(word) - 1):
 
        # Return true as word is found
        return True
 
    temp = board[row][col]
 
    # Mark cell visited
    board[row][col] = '*'
 
    # Check Adjacent cells
    # for the next character
    if (checkWord(board, word,
                  index + 1,
                  row + 1, col) or
        checkWord(board, word,
                  index + 1,
                  row - 1, col) or
        checkWord(board, word,
                  index + 1,
                  row, col + 1) or
        checkWord(board, word,
                  index + 1,
                  row, col - 1)):
        board[row][col] = temp
        return True
    
    # Restore cell value
    board[row][col] = temp
    return False
 
# Driver Code
if __name__ == "__main__":
 
    board = [['A', 'B', 'C', 'E'],
            ['S', 'F', 'C', 'S'],
            ['A', 'D', 'E', 'E']]
    word = "CFDASABCESEE"
    f = 0
     
    for i in range (len(board)):
        for j in range (len(board[0])):
            if (board[i][j] == word[0] and
                checkWord(board, word,
                          0, i, j)):
                print ("True" )
                f = 1
                break
        if f == 1:
          break
           
    if f == 0:
       print ("False")
 
# This code is contributed by Chitranayal

C#




// C# program to check if a given word
// can be formed from the adjacent
// characters in a matrix of characters
using System;
 
class GFG{
 
// Function to check if the word exists
static bool checkWord(char [,]board, String word,
                      int index, int row, int col)
{
     
    // If index exceeds board range
    if (row < 0 || col < 0 ||
        row >= board.GetLength(0) ||
        col >= board.GetLength(1))
        return false;
 
    // If the current cell does not
    // contain the required character
    if (board[row, col] != word[index])
        return false;
 
    // If the cell contains the required
    // character and is the last character
    // of the word required to be matched
    else if (index == word.Length - 1)
 
        // Return true as word is found
        return true;
 
    char temp = board[row, col];
 
    // Mark cell visited
    board[row, col] = '*';
 
    // Check adjacent cells
    // for the next character
    if (checkWord(board, word,
                  index + 1, row + 1, col) ||
        checkWord(board, word,
                  index + 1, row - 1, col) ||
        checkWord(board, word,
                  index + 1, row, col + 1) ||
        checkWord(board, word,
                  index + 1, row, col - 1))
    {
        board[row, col] = temp;
 
        return true;
    }
 
    // Restore cell value
    board[row, col] = temp;
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    char[,] board = { { 'A', 'B', 'C', 'E' },
                      { 'S', 'F', 'C', 'S' },
                      { 'A', 'D', 'E', 'E' } };
    String word = "CFDASABCESEE";
 
    for(int i = 0; i < board.GetLength(0); i++)
    {
       for(int j = 0; j < board.GetLength(1); j++)
       {
          if (board[i, j] == word[0] &&
              checkWord(board, word, 0, i, j))
          {
              Console.WriteLine("True");
              return;
          }
       }
    }
    Console.WriteLine("False");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript program to check if a given word
// can be formed from the adjacent
// characters in a matrix of characters
 
// Function to check if the word exists
function checkWord(board, word, index, row, col)
{
     
    // If index exceeds board range
    if (row < 0 || col < 0 ||
        row >= board.length ||
        col >= board[0].length)
        return false;
     
    // If the current cell does not
    // contain the required character
    if (board[row][col] !== word[index])
        return false;
         
    // If the cell contains the required
    // character and is the last character
    // of the word required to be matched
    else if (index === word.length - 1)
     
        // Return true as word is found
        return true;
     
    var temp = board[row][col];
     
    // Mark cell visited
    board[row][col] = "*";
     
    // Check adjacent cells
    // for the next character
    if (checkWord(board, word,
                  index + 1, row + 1, col) ||
        checkWord(board, word,
                  index + 1, row - 1, col) ||
        checkWord(board, word,
                  index + 1, row, col + 1) ||
        checkWord(board, word,
                  index + 1, row, col - 1))
    {
        board[row][col] = temp;
        return true;
    }
     
    // Restore cell value
    board[row][col] = temp;
    return false;
}
 
// Driver Code
var board = [ [ "A", "B", "C", "E" ],
              [ "S", "F", "C", "S" ],
              [ "A", "D", "E", "E" ],];
var word = "CFDASABCESEE";
var f = 0;
 
for(var i = 0; i < board.length; i++)
{
    for(var j = 0; j < board[0].length; j++)
    {
        if (board[i][j] === word[0] &&
            checkWord(board, word, 0, i, j))
        {
            document.write("True");
            f = 1;
        }
    }
    if (f === 1)
    {
        i = board.length + 1;
    }
}
if (f === 0)
{
    document.write("False");
}
 
// This code is contributed by rdtank
 
</script>
Output: 
True

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :