Find length of the largest region in Boolean Matrix

Consider a matrix with rows and columns, where each cell contains either a ‘0’ or a ‘1’ and any cell containing a 1 is called a filled cell. Two cells are said to be connected if they are adjacent to each other horizontally, vertically, or diagonally. If one or more filled cells are also connected, they form a region. find the length of the largest region.

Examples:

Input : M[][5] = { 0 0 1 1 0
                   1 0 1 1 0
                   0 1 0 0 0
                   0 0 0 0 1 }
Output : 6 
In the following example, there are 
2 regions one with length 1 and the 
other as 6. So largest region: 6

Input : M[][5] = { 0 0 1 1 0
                   0 0 1 1 0
                   0 0 0 0 0
                   0 0 0 0 1 }
Output: 6 
In the following example, there are 
2 regions one with length 1 and the 
other as 4. So largest region: 4

Asked in: Amazon interview

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Solution: The idea is based on the problem or finding number of islands in Boolean 2D-matrix
Approach:

A cell in 2D matrix can be connected to at most 8 neighbours.
So in DFS, make recursive calls for 8 neighbours of that cell.
Keep a visited Hash-map to keep track of all visited cells.
Also keep track of the visited 1’s in every DFS and update maximum length region.

Below is implementation of above idea.

C++

// Program to find the length of the largest 
// region in boolean 2D-matrix 
#include <bits/stdc++.h> 
using namespace std; 
#define ROW 4 
#define COL 5 
  
// A function to check if 
// a given cell (row, col) 
// can be included in DFS 
int isSafe(int M[][COL], int row, 
           int col, 
           bool visited[][COL]) 
{ 
    // row number is in range, 
    // column number is in 
    // range and value is 1 
    // and not yet visited 
    return (row >= 0) 
           && (row < ROW) 
           && (col >= 0) 
           && (col < COL) 
           && (M[row][col] 
               && !visited[row][col]); 
} 
  
// A utility function to 
// do DFS for a 2D boolean 
// matrix. It only considers 
// the 8 neighbours as 
// adjacent vertices 
void DFS(int M[][COL], int row, int col, 
         bool visited[][COL], int& count) 
{ 
    // These arrays are used 
    // to get row and column 
    // numbers of 8 neighbours 
    // of a given cell 
    static int rowNbr[] = { -1, -1, -1, 0, 0, 1, 1, 1 }; 
    static int colNbr[] = { -1, 0, 1, -1, 1, -1, 0, 1 }; 
  
    // Mark this cell as visited 
    visited[row][col] = true; 
  
    // Recur for all connected neighbours 
    for (int k = 0; k < 8; ++k) { 
        if (isSafe(M, row + rowNbr[k], col + colNbr[k], 
                   visited)) { 
            // Increment region length by one 
            count++; 
            DFS(M, row + rowNbr[k], col + colNbr[k], 
                visited, count); 
        } 
    } 
} 
  
// The main function that returns 
// largest  length region 
// of a given boolean 2D matrix 
int largestRegion(int M[][COL]) 
{ 
    // Make a bool array to mark visited cells. 
    // Initially all cells are unvisited 
    bool visited[ROW][COL]; 
    memset(visited, 0, sizeof(visited)); 
  
    // Initialize result as 0 and travesle through the 
    // all cells of given matrix 
    int result = INT_MIN; 
    for (int i = 0; i < ROW; ++i) { 
        for (int j = 0; j < COL; ++j) { 
            // If a cell with value 1 is not 
            if (M[i][j] && !visited[i][j]) { 
                // visited yet, then new region found 
                int count = 1; 
                DFS(M, i, j, visited, count); 
  
                // maximum region 
                result = max(result, count); 
            } 
        } 
    } 
    return result; 
} 
  
// Driver program to test above function 
int main() 
{ 
    int M[][COL] = { { 0, 0, 1, 1, 0 }, 
                     { 1, 0, 1, 1, 0 }, 
                     { 0, 1, 0, 0, 0 }, 
                     { 0, 0, 0, 0, 1 } }; 
  
    cout << largestRegion(M); 
  
    return 0; 
} 

Java

// Java program to find the length of the largest 
// region in boolean 2D-matrix 
import java.io.*; 
import java.util.*; 
  
class GFG { 
    static int ROW, COL, count; 
  
    // A function to check if a given cell (row, col) 
    // can be included in DFS 
    static boolean isSafe(int[][] M, int row, 
                          int col, boolean[][] visited) 
    { 
        // row number is in range, column number is in 
        // range and value is 1 and not yet visited 
        return ((row >= 0) && (row < ROW) && (col >= 0) 
                && (col < COL) && (M[row][col] == 1 && !visited[row][col])); 
    } 
  
    // A utility function to do DFS for a 2D boolean 
    // matrix. It only considers the 8 neighbours as 
    // adjacent vertices 
    static void DFS(int[][] M, int row, 
                    int col, boolean[][] visited) 
    { 
        // These arrays are used to get row and column 
        // numbers of 8 neighbours of a given cell 
        int[] rowNbr = { -1, -1, -1, 0, 0, 1, 1, 1 }; 
        int[] colNbr = { -1, 0, 1, -1, 1, -1, 0, 1 }; 
  
        // Mark this cell as visited 
        visited[row][col] = true; 
  
        // Recur for all connected neighbours 
        for (int k = 0; k < 8; k++) { 
            if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited)) { 
                // increment region length by one 
                count++; 
                DFS(M, row + rowNbr[k], col + colNbr[k], visited); 
            } 
        } 
    } 
  
    // The main function that returns largest length region 
    // of a given boolean 2D matrix 
    static int largestRegion(int[][] M) 
    { 
        // Make a boolean array to mark visited cells. 
        // Initially all cells are unvisited 
        boolean[][] visited = new boolean[ROW][COL]; 
  
        // Initialize result as 0 and traverse through the 
        // all cells of given matrix 
        int result = 0; 
        for (int i = 0; i < ROW; i++) { 
            for (int j = 0; j < COL; j++) { 
  
                // If a cell with value 1 is not 
                if (M[i][j] == 1 && !visited[i][j]) { 
  
                    // visited yet, then new region found 
                    count = 1; 
                    DFS(M, i, j, visited); 
  
                    // maximum region 
                    result = Math.max(result, count); 
                } 
            } 
        } 
        return result; 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int M[][] = { { 0, 0, 1, 1, 0 }, 
                      { 1, 0, 1, 1, 0 }, 
                      { 0, 1, 0, 0, 0 }, 
                      { 0, 0, 0, 0, 1 } }; 
        ROW = 4; 
        COL = 5; 
        System.out.println(largestRegion(M)); 
    } 
} 
  
// This code is contributed by rachana soma 

Python3

# Python3 program to find the length of the  
# largest region in boolean 2D-matrix  
  
# A function to check if a given cell  
# (row, col) can be included in DFS  
def isSafe(M, row, col, visited): 
    global ROW, COL 
      
    # row number is in range, column number is in  
    # range and value is 1 and not yet visited  
    return ((row >= 0) and (row < ROW) and
            (col >= 0) and (col < COL) and 
            (M[row][col] and not visited[row][col])) 
  
# A utility function to do DFS for a 2D  
# boolean matrix. It only considers  
# the 8 neighbours as adjacent vertices  
def DFS(M, row, col, visited, count): 
      
    # These arrays are used to get row and column  
    # numbers of 8 neighbours of a given cell  
    rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1]  
    colNbr = [-1, 0, 1, -1, 1, -1, 0, 1]  
  
    # Mark this cell as visited  
    visited[row][col] = True
  
    # Recur for all connected neighbours  
    for k in range(8): 
        if (isSafe(M, row + rowNbr[k],  
                   col + colNbr[k], visited)): 
                         
            # increment region length by one  
            count[0] += 1
            DFS(M, row + rowNbr[k],  
                col + colNbr[k], visited, count) 
  
# The main function that returns largest 
# length region of a given boolean 2D matrix  
def largestRegion(M): 
    global ROW, COL 
      
    # Make a bool array to mark visited cells.  
    # Initially all cells are unvisited  
    visited = [[0] * COL for i in range(ROW)] 
  
    # Initialize result as 0 and travesle  
    # through the all cells of given matrix  
    result = -999999999999
    for i in range(ROW): 
        for j in range(COL): 
              
            # If a cell with value 1 is not  
            if (M[i][j] and not visited[i][j]): 
                  
                # visited yet, then new region found  
                count = [1]  
                DFS(M, i, j, visited, count)  
  
                # maximum region  
                result = max(result, count[0]) 
    return result 
  
# Driver Code 
ROW = 4
COL = 5
  
M = [[0, 0, 1, 1, 0], 
     [1, 0, 1, 1, 0],  
     [0, 1, 0, 0, 0], 
     [0, 0, 0, 0, 1]]  
  
print(largestRegion(M)) 
  
# This code is contributed by PranchalK 

C#

// C# program to find the length of 
// the largest region in boolean 2D-matrix 
using System; 
  
class GFG { 
    static int ROW, COL, count; 
  
    // A function to check if a given cell 
    // (row, col) can be included in DFS 
    static Boolean isSafe(int[, ] M, int row, 
                          int col, 
                          Boolean[, ] visited) 
    { 
        // row number is in range, column number is in 
        // range and value is 1 and not yet visited 
        return ((row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row, col] == 1 && !visited[row, col])); 
    } 
  
    // A utility function to do DFS for a 2D boolean 
    // matrix. It only considers the 8 neighbours as 
    // adjacent vertices 
    static void DFS(int[, ] M, int row, 
                    int col, Boolean[, ] visited) 
    { 
        // These arrays are used to get row and column 
        // numbers of 8 neighbours of a given cell 
        int[] rowNbr = { -1, -1, -1, 0, 0, 1, 1, 1 }; 
        int[] colNbr = { -1, 0, 1, -1, 1, -1, 0, 1 }; 
  
        // Mark this cell as visited 
        visited[row, col] = true; 
  
        // Recur for all connected neighbours 
        for (int k = 0; k < 8; k++) { 
            if (isSafe(M, row + rowNbr[k], 
                       col + colNbr[k], visited)) { 
                // increment region length by one 
                count++; 
                DFS(M, row + rowNbr[k], col + colNbr[k], visited); 
            } 
        } 
    } 
  
    // The main function that returns 
    // largest length region of 
    // a given boolean 2D matrix 
    static int largestRegion(int[, ] M) 
    { 
        // Make a boolean array to mark visited cells. 
        // Initially all cells are unvisited 
        Boolean[, ] visited = new Boolean[ROW, COL]; 
  
        // Initialize result as 0 and traverse 
        // through the all cells of given matrix 
        int result = 0; 
        for (int i = 0; i < ROW; i++) { 
            for (int j = 0; j < COL; j++) { 
  
                // If a cell with value 1 is not 
                if (M[i, j] == 1 && !visited[i, j]) { 
  
                    // visited yet, 
                    // then new region found 
                    count = 1; 
                    DFS(M, i, j, visited); 
  
                    // maximum region 
                    result = Math.Max(result, count); 
                } 
            } 
        } 
        return result; 
    } 
  
    // Driver code 
    public static void Main(String[] args) 
    { 
        int[, ] M = { { 0, 0, 1, 1, 0 }, 
                      { 1, 0, 1, 1, 0 }, 
                      { 0, 1, 0, 0, 0 }, 
                      { 0, 0, 0, 0, 1 } }; 
        ROW = 4; 
        COL = 5; 
        Console.WriteLine(largestRegion(M)); 
    } 
} 
  
// This code is contributed by Rajput-Ji 

Output:

Complexity Analysis:

Time complexity: O(ROW * COL).
At worst case all the cells will be visited so time complexity is O(ROW * COL).
Auxiliary Space:O(ROW * COL).
To store the visited nodes O(ROW * COL) space is needed.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

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