Write a C# 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
-
If M[i][j] is 1 then
- 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:
// C# Code for Maximum size square // sub-matrix with all 1s using System;
public class GFG {
// method for Maximum size square sub-matrix with all 1s
static void printMaxSubSquare( int [, ] M)
{
int i, j;
// no of rows in M[,]
int R = M.GetLength(0);
// no of columns in M[,]
int C = M.GetLength(1);
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;
}
}
}
Console.WriteLine( "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--) {
Console.Write(M[i, j] + " " );
}
Console.WriteLine();
}
}
// Driver program
public static void Main()
{
int [, ] M = new int [6, 5] {
{ 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);
}
} |
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.
C# 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:
// C# code to implement the approach using System;
using System.Numerics;
using System.Collections.Generic;
public class GFG {
static int R = 6;
static int C = 5;
static void printMaxSubSquare( int [, ] M)
{
int [, ] S = new int [2, C];
int Maxx = 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
Maxx = Math.Max(Maxx, Entrie);
}
}
// Print the square
Console.Write( "Maximum size sub-matrix is: \n" );
for ( int i = 0; i < Maxx; i++) {
for ( int j = 0; j < Maxx; j++) {
Console.Write( "1 " );
}
Console.WriteLine();
}
}
// 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);
}
} |
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!