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Java Program for Maximum size square sub-matrix with all 1s

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Write a Java program for a given binary matrix, the task is to find out the maximum size square sub-matrix with all 1s.

Approach:

Let the given binary matrix be M[R][C]. The idea of the algorithm is to construct an auxiliary size matrix S[][] in which each entry S[i][j] represents the size of the square sub-matrix with all 1s including M[i][j] where M[i][j] is the rightmost and bottom-most entry in sub-matrix.

Step-by-step approach:

  • Construct a sum matrix S[R][C] for the given M[R][C].
    • Copy first row and first columns as it is from M[][] to S[][]
    • For other entries, use the following expressions to construct S[][]
      • If M[i][j] is 1 then
        • S[i][j] = min(S[i][j-1], S[i-1][j], S[i-1][j-1]) + 1
      • Else If M[i][j] is 0 then
        • S[i][j] = 0
  • Find the maximum entry in S[R][C]
  • Using the value and coordinates of maximum entry in S[i], print sub-matrix of M[][]

Below is the implementation of the above approach:

Java

// JAVA Code for Maximum size square
// sub-matrix with all 1s
public class GFG {
    // method for Maximum size square sub-matrix with all 1s
    static void printMaxSubSquare(int M[][])
    {
        int i, j;
        int R = M.length; // no of rows in M[][]
        int C = M[0].length; // no of columns in M[][]
        int S[][] = new int[R][C];
 
        int max_of_s, max_i, max_j;
 
        /* Set first column of S[][]*/
        for (i = 0; i < R; i++)
            S[i][0] = M[i][0];
 
        /* Set first row of S[][]*/
        for (j = 0; j < C; j++)
            S[0][j] = M[0][j];
 
        /* Construct other entries of S[][]*/
        for (i = 1; i < R; i++) {
            for (j = 1; j < C; j++) {
                if (M[i][j] == 1)
                    S[i][j] = Math.min(
                                S[i][j - 1],
                                Math.min(S[i - 1][j],
                                        S[i - 1][j - 1]))
                            + 1;
                else
                    S[i][j] = 0;
            }
        }
 
        /* Find the maximum entry, and indexes of maximum
        entry in S[][] */
        max_of_s = S[0][0];
        max_i = 0;
        max_j = 0;
        for (i = 0; i < R; i++) {
            for (j = 0; j < C; j++) {
                if (max_of_s < S[i][j]) {
                    max_of_s = S[i][j];
                    max_i = i;
                    max_j = j;
                }
            }
        }
 
        System.out.println("Maximum size sub-matrix is: ");
        for (i = max_i; i > max_i - max_of_s; i--) {
            for (j = max_j; j > max_j - max_of_s; j--) {
                System.out.print(M[i][j] + " ");
            }
            System.out.println();
        }
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int M[][]
            = { { 0, 1, 1, 0, 1 }, { 1, 1, 0, 1, 0 },
                { 0, 1, 1, 1, 0 }, { 1, 1, 1, 1, 0 },
                { 1, 1, 1, 1, 1 }, { 0, 0, 0, 0, 0 } };
 
        printMaxSubSquare(M);
    }
}

                    

Output
Maximum size sub-matrix is: 
1 1 1 
1 1 1 
1 1 1 

Time Complexity: O(m*n), where m is the number of rows and n is the number of columns in the given matrix.
Auxiliary Space: O(m*n), where m is the number of rows and n is the number of columns in the given matrix.

Java Program for Maximum size square sub-matrix with all 1s using Dynamic Programming:

In order to compute an entry at any position in the matrix we only need the current row and the previous row.

Below is the implementation of the above approach:

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    static int R = 6;
    static int C = 5;
 
    static void printMaxSubSquare(int M[][])
    {
        int S[][] = new int[2][C], Max = 0;
 
        // set all elements of S to 0 first
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < C; j++) {
                S[i][j] = 0;
            }
        }
 
        // Construct the entries
        for (int i = 0; i < R; i++) {
            for (int j = 0; j < C; j++) {
 
                // Compute the entrie at the current
                // position
                int Entrie = M[i][j];
                if (Entrie != 0) {
                    if (j != 0) {
                        Entrie = 1
                                + Math.min(
                                    S[1][j - 1],
                                    Math.min(S[0][j - 1],
                                            S[1][j]));
                    }
                }
 
                // Save the last entrie and add the new one
                S[0][j] = S[1][j];
                S[1][j] = Entrie;
 
                // Keep track of the max square length
                Max = Math.max(Max, Entrie);
            }
        }
 
        // Print the square
        System.out.print("Maximum size sub-matrix is: \n");
        for (int i = 0; i < Max; i++) {
            for (int j = 0; j < Max; j++) {
                System.out.print("1 ");
            }
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int M[][]
            = { { 0, 1, 1, 0, 1 }, { 1, 1, 0, 1, 0 },
                { 0, 1, 1, 1, 0 }, { 1, 1, 1, 1, 0 },
                { 1, 1, 1, 1, 1 }, { 0, 0, 0, 0, 0 } };
 
        printMaxSubSquare(M);
    }
}
 
// This code is contributed by code_hunt.

                    

Output
Maximum size sub-matrix is: 
1 1 1 
1 1 1 
1 1 1 

Time Complexity: O(m*n) where m is the number of rows and n is the number of columns in the given matrix.
Auxiliary space: O(n) where n is the number of columns in the given matrix.

Please refer complete article on Maximum size square sub-matrix with all 1s for more details!



Last Updated : 24 Oct, 2023
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