Find length of the largest region in Boolean Matrix

Difficulty Level : Medium
Last Updated : 06 Jul, 2021

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: 4
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

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 code
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 } };
 
    // Function call
    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;
       
        // Function call
        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]]
 
# Function call
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;
       
        // Function call
        Console.WriteLine(largestRegion(M));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript program to find the length of the largest
// region in boolean 2D-matrix
 
let ROW, COL, count;
 
// A function to check if a given cell (row, col)
    // can be included in DFS
function isSafe(M,row,col,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
function DFS(M,row,col,visited)
{
    // These arrays are used to get row and column
        // numbers of 8 neighbours of a given cell
        let rowNbr = [ -1, -1, -1, 0, 0, 1, 1, 1 ];
        let colNbr = [ -1, 0, 1, -1, 1, -1, 0, 1 ];
  
        // Mark this cell as visited
        visited[row][col] = true;
  
        // Recur for all connected neighbours
        for (let 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
function largestRegion(M)
{
    // Make a boolean array to mark visited cells.
        // Initially all cells are unvisited
        let visited = new Array(ROW);
         for(let i=0;i<ROW;i++)
        {
            visited[i]=new Array(COL);
            for(let j=0;j<COL;j++)
            {
                visited[i][j]=false;
            }
        }
         
         
        // Initialize result as 0 and traverse through the
        // all cells of given matrix
        let result = 0;
        for (let i = 0; i < ROW; i++)
        {
            for (let 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
let 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;
 
// Function call
document.write(largestRegion(M));
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>

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.

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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