Search a string in Matrix Using Split function in Java

Given a string str and a matrix mat[][] of lowercase English alphabets, the task is to find whether the string str appears in the matrix (either row-wise or column-wise).

Examples:

Input: str = "GFG"
            mat[][] = {{'G', 'E', 'E', 'K', 'S'}, 
                             {'F', 'O', 'R', 'A', 'N'}, 
                             {'G', 'E', 'E', 'K', 'S'}}
Output: Yes
GFG is present in the first column.

Input: str = "SSS"
            mat[][] = {{'G', 'E', 'E', 'K', 'S'}, 
                              {'F', 'O', 'R', 'A', 'N'}, 
                              {'G', 'E', 'E', 'K', 'S'}}
Output: No

Approach:
If the search string is present in any row of the matrix the following results will be produced by the split function:

  1. If the string occupies the whole row then the split function will return an array of length zero.
  2. If the string is present in between characters then the array length will be greater than one.
  3. If the array length is one then three possible cases can be there-
    • The string occurs in first half.
    • The string occurs in second half.
    • The string is not present in that row.
  4. To search the string column wise transpose the matrix and repeat step one.
  5. Print Yes if the string is found else print No.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to search a string in
// the matrix using split function
import java.util.*;
public class GFG {
  
    // Function to check the occurrence of string in the matrix
    public static int check(String[] matrix, String string)
    {
        // Looping the contents in the matrix
        for (String input : matrix) {
  
            // using split operator
            String[] value = input.split(string);
  
            if (value.length >= 2 || value.length == 0) {
                return 1;
            }
            else if (value.length == 1
                     && input.length() != value[0].length()) {
                return 1;
            }
        }
        return 0;
    }
  
    // Function to transpose the given matrix
    public static String[] vertical(String[] matrix)
    {
        String[] vertical_value = new String[matrix[0].length()];
        for (int i = 0; i < matrix[0].length(); i++) {
            String temp = "";
            for (int j = 0; j < matrix.length; j++)
                temp += matrix[j].charAt(i);
            vertical_value[i] = temp;
        }
  
        // returning the transposed matrix
        return vertical_value;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        // Input matrix of characters
        String[] matrix = { "GEEKS", "FORAN", "GEEKS" };
  
        // element to be searched
        String search = "GFG";
  
        // Transpose of the matrix
        String[] verticalMatrix = vertical(matrix);
  
        // Searching process
        int horizontal_search = check(matrix, search);
        int vertical_search = check(verticalMatrix, search);
  
        // If found
        if (horizontal_search == 1 || vertical_search == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

chevron_right


Output:

Yes

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.




My Personal Notes arrow_drop_up


If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.