Find the number of islands | Set 1 (Using DFS)

Given a boolean 2D matrix, find the number of islands. A group of connected 1s forms an island. For example, the below matrix contains 5 islands

Example:

Input : mat[][] = {{1, 1, 0, 0, 0},
                   {0, 1, 0, 0, 1},
                   {1, 0, 0, 1, 1},
                   {0, 0, 0, 0, 0},
                   {1, 0, 1, 0, 1}}
Output : 5

This is a variation of the standard problem: “Counting the number of connected components in an undirected graph”. 

Before we go to the problem, let us understand what is a connected component. A connected component of an undirected graph is a subgraph in which every two vertices are connected to each other by a path(s), and which is connected to no other vertices outside the subgraph. 
For example, the graph shown below has three connected components. 

A graph where all vertices are connected with each other has exactly one connected component, consisting of the whole graph. Such a graph with only one connected component is called a Strongly Connected Graph.
The problem can be easily solved by applying DFS() on each component. In each DFS() call, a component or a sub-graph is visited. We will call DFS on the next un-visited component. The number of calls to DFS() gives the number of connected components. BFS can also be used.

What is an island?

A group of connected 1s forms an island. For example, the below matrix contains 4 islands

A cell in 2D matrix can be connected to 8 neighbours. So, unlike standard DFS(), where we recursively call for all adjacent vertices, here we can recursively call for 8 neighbours only. We keep track of the visited 1s so that they are not visited again. 

C++

// C++ Program to count islands in boolean 2D matrix
#include <bits/stdc++.h>
using namespace std;

#define ROW 5
#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])
{
    // 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))
            DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}

// The main function that returns
// count of islands in a given boolean
// 2D matrix
int countIslands(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 count as 0 and
    // traverse through the all cells of
    // given matrix
    int count = 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] && !visited[i][j]) {
                // visited yet, then new island found
                // Visit all cells in this island.
                DFS(M, i, j, visited);

                // and increment island count
                ++count;
            }

    return count;
}

// Driver code
int main()
{
    int M[][COL] = { { 1, 1, 0, 0, 0 },
                     { 0, 1, 0, 0, 1 },
                     { 1, 0, 0, 1, 1 },
                     { 0, 0, 0, 0, 0 },
                     { 1, 0, 1, 0, 1 } };

    cout << "Number of islands is: " << countIslands(M);

    return 0;
}

// This is code is contributed by rathbhupendra

C

// Program to count islands in boolean 2D matrix
#include <stdbool.h>
#include <stdio.h>
#include <string.h>

#define ROW 5
#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])
{
    // 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))
            DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}

// The main function that returns count of islands in a given boolean
// 2D matrix
int countIslands(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 count as 0 and traverse through the all cells of
    // given matrix
    int count = 0;
    for (int i = 0; i < ROW; ++i)
        for (int j = 0; j < COL; ++j)
            if (M[i][j] && !visited[i][j]) // If a cell with value 1 is not
            { // visited yet, then new island found
                DFS(M, i, j, visited); // Visit all cells in this island.
                ++count; // and increment island count
            }

    return count;
}

// Driver program to test above function
int main()
{
    int M[][COL] = { { 1, 1, 0, 0, 0 },
                     { 0, 1, 0, 0, 1 },
                     { 1, 0, 0, 1, 1 },
                     { 0, 0, 0, 0, 0 },
                     { 1, 0, 1, 0, 1 } };

    printf("Number of islands is: %d\n", countIslands(M));

    return 0;
}

Java

// Java program to count islands in boolean 2D matrix
import java.util.*;
import java.lang.*;
import java.io.*;

class Islands {
    // No of rows and columns
    static final int ROW = 5, COL = 5;

    // A function to check if a given cell (row, col) can
    // be included in DFS
    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 neighbors as adjacent vertices
    void DFS(int M[][], int row, int col, boolean visited[][])
    {
        // These arrays are used to get row and column numbers
        // of 8 neighbors of a given cell
        int rowNbr[] = new int[] { -1, -1, -1, 0, 0, 1, 1, 1 };
        int colNbr[] = new int[] { -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))
                DFS(M, row + rowNbr[k], col + colNbr[k], visited);
    }

    // The main function that returns count of islands in a given
    // boolean 2D matrix
    int countIslands(int M[][])
    {
        // Make a bool array to mark visited cells.
        // Initially all cells are unvisited
        boolean visited[][] = new boolean[ROW][COL];

        // Initialize count as 0 and traverse through the all cells
        // of given matrix
        int count = 0;
        for (int i = 0; i < ROW; ++i)
            for (int j = 0; j < COL; ++j)
                if (M[i][j] == 1 && !visited[i][j]) // If a cell with
                { // value 1 is not
                    // visited yet, then new island found, Visit all
                    // cells in this island and increment island count
                    DFS(M, i, j, visited);
                    ++count;
                }

        return count;
    }

    // Driver method
    public static void main(String[] args) throws java.lang.Exception
    {
        int M[][] = new int[][] { { 1, 1, 0, 0, 0 },
                                  { 0, 1, 0, 0, 1 },
                                  { 1, 0, 0, 1, 1 },
                                  { 0, 0, 0, 0, 0 },
                                  { 1, 0, 1, 0, 1 } };
        Islands I = new Islands();
        System.out.println("Number of islands is: " + I.countIslands(M));
    }
} // Contributed by Aakash Hasija

Python3

# Program to count islands in boolean 2D matrix
class Graph:

    def __init__(self, row, col, g):
        self.ROW = row
        self.COL = col
        self.graph = g

    # A function to check if a given cell
    # (row, col) can be included in DFS
    def isSafe(self, i, j, visited):
        # row number is in range, column number
        # is in range and value is 1
        # and not yet visited
        return (i >= 0 and i < self.ROW and
                j >= 0 and j < self.COL and
                not visited[i][j] and self.graph[i][j])


    # A utility function to do DFS for a 2D
    # boolean matrix. It only considers
    # the 8 neighbours as adjacent vertices
    def DFS(self, i, j, visited):

        # 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[i][j] = True

        # Recur for all connected neighbours
        for k in range(8):
            if self.isSafe(i + rowNbr[k], j + colNbr[k], visited):
                self.DFS(i + rowNbr[k], j + colNbr[k], visited)


    # The main function that returns
    # count of islands in a given boolean
    # 2D matrix
    def countIslands(self):
        # Make a bool array to mark visited cells.
        # Initially all cells are unvisited
        visited = [[False for j in range(self.COL)]for i in range(self.ROW)]

        # Initialize count as 0 and traverse
        # through the all cells of
        # given matrix
        count = 0
        for i in range(self.ROW):
            for j in range(self.COL):
                # If a cell with value 1 is not visited yet,
                # then new island found
                if visited[i][j] == False and self.graph[i][j] == 1:
                    # Visit all cells in this island
                    # and increment island count
                    self.DFS(i, j, visited)
                    count += 1

        return count


graph = [[1, 1, 0, 0, 0],
        [0, 1, 0, 0, 1],
        [1, 0, 0, 1, 1],
        [0, 0, 0, 0, 0],
        [1, 0, 1, 0, 1]]


row = len(graph)
col = len(graph[0])

g = Graph(row, col, graph)

print ("Number of islands is:")
print (g.countIslands())

# This code is contributed by Neelam Yadav

C#

// C# program to count
// islands in boolean
// 2D matrix
using System;

class GFG {
    // No of rows
    // and columns
    static int ROW = 5, COL = 5;

    // A function to check if
    // a given cell (row, col)
    // can be included in DFS
    static bool isSafe(int[, ] M, int row,
                       int col, bool[, ] 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
    // neighbors as adjacent vertices
    static void DFS(int[, ] M, int row,
                    int col, bool[, ] visited)
    {
        // These arrays are used to
        // get row and column numbers
        // of 8 neighbors of a given cell
        int[] rowNbr = new int[] { -1, -1, -1, 0,
                                   0, 1, 1, 1 };
        int[] colNbr = new int[] { -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))
                DFS(M, row + rowNbr[k],
                    col + colNbr[k], visited);
    }

    // The main function that
    // returns count of islands
    // in a given boolean 2D matrix
    static int countIslands(int[, ] M)
    {
        // Make a bool array to
        // mark visited cells.
        // Initially all cells
        // are unvisited
        bool[, ] visited = new bool[ROW, COL];

        // Initialize count as 0 and
        // traverse through the all
        // cells of given matrix
        int count = 0;
        for (int i = 0; i < ROW; ++i)
            for (int j = 0; j < COL; ++j)
                if (M[i, j] == 1 && !visited[i, j]) {
                    // If a cell with value 1 is not
                    // visited yet, then new island
                    // found, Visit all cells in this
                    // island and increment island count
                    DFS(M, i, j, visited);
                    ++count;
                }

        return count;
    }

    // Driver Code
    public static void Main()
    {
        int[, ] M = new int[, ] { { 1, 1, 0, 0, 0 },
                                  { 0, 1, 0, 0, 1 },
                                  { 1, 0, 0, 1, 1 },
                                  { 0, 0, 0, 0, 0 },
                                  { 1, 0, 1, 0, 1 } };
        Console.Write("Number of islands is: " + countIslands(M));
    }
}

// This code is contributed
// by shiv_bhakt.

PHP

<?php
// Program to count islands
// in boolean 2D matrix

$ROW = 5;
$COL = 5;

// A function to check if a
// given cell (row, col) can
// be included in DFS
function 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 ($row >= 0) && ($row < $ROW) &&
           ($col >= 0) && ($col < $COL) &&
           ($M[$row][$col] &&
             !isset($visited[$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
    $rowNbr = array(-1, -1, -1, 0,
                    0, 1, 1, 1);
    $colNbr = array(-1, 0, 1, -1,
                    1, -1, 0, 1);

    // Mark this cell as visited
    $visited[$row][$col] = true;

    // Recur for all
    // connected neighbours
    for ($k = 0; $k < 8; ++$k)
        if (isSafe($M, $row + $rowNbr[$k],
                $col + $colNbr[$k], $visited))
            DFS($M, $row + $rowNbr[$k],
                $col + $colNbr[$k], $visited);
}

// The main function that returns
// count of islands in a given
// boolean 2D matrix
function countIslands(&$M)
{
    global $ROW, $COL;

    // Make a bool array to
    // mark visited cells.
    // Initially all cells
    // are unvisited
    $visited = array(array());

    // Initialize count as 0 and
    // traverse through the all
    // cells of given matrix
    $count = 0;
    for ($i = 0; $i < $ROW; ++$i)
        for ($j = 0; $j < $COL; ++$j)
            if ($M[$i][$j] &&
                 !isset($visited[$i][$j])) // If a cell with value 1
            {                               // is not visited yet,
                DFS($M, $i, $j, $visited); // then new island found
                ++$count;                   // Visit all cells in this
            }                               // island and increment
                                           // island count.

    return $count;
}

// Driver Code
$M = array(array(1, 1, 0, 0, 0),
           array(0, 1, 0, 0, 1),
           array(1, 0, 0, 1, 1),
            array(0, 0, 0, 0, 0),
           array(1, 0, 1, 0, 1));

echo "Number of islands is: ",
            countIslands($M);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// Javascript program to count islands in boolean 2D matrix

    // No of rows and columns
    let  ROW = 5, COL = 5;

    // 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 neighbors as adjacent vertices
    function DFS(M, row, col, visited)
    {
        // These arrays are used to get row and column numbers
        // of 8 neighbors 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))
            {
                DFS(M, row + rowNbr[k], col + colNbr[k], visited);
            }
        }

    }

    // The main function that returns count of islands in a given
    // boolean 2D matrix
    function countIslands(M)
    {
        // Make a bool 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 i = 0; i < ROW; i++)
        {
            for(let j = 0; j < COL; j++)
            {
                visited[i][j] = false;
            }
        }
        // Initialize count as 0 and traverse through the all cells
        // of given matrix
        let count = 0;
        for (let i = 0; i < ROW; ++i)
        {
            for (let j = 0; j < COL; ++j)
            {
                if (M[i][j] == 1 && !visited[i][j])
                {
                    // value 1 is not
                    // visited yet, then new island found, Visit all
                    // cells in this island and increment island count
                    DFS(M, i, j, visited);
                    count++;
                }
            }
        }
        return count;
    }

    // Driver method
    let M = [[ 1, 1, 0, 0, 0 ], [ 0, 1, 0, 0, 1],
    [1, 0, 0, 1, 1] ,[0, 0, 0, 0, 0], [1, 0, 1, 0, 1]];
    document.write("Number of islands is: " + countIslands(M));

    // This code is contributed by avanitrachhadiya2155
</script>

Output

Number of islands is: 5

Time complexity: O(ROW x COL)
Auxiliary Space: O(ROW x COL), due to visited matrix

Alternative solution without creating visited matrix:

C++

// C++Program to count islands in boolean 2D matrix
#include <bits/stdc++.h>
using namespace std;

// A utility function to do DFS for a 2D
//  boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(vector<vector<int>> &M, int i, int j, int ROW,
         int COL)
{
    //Base condition
    //if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL-  or if M[i][j] != 1 then we will simply return
    if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
    {
        return;
    }

    if (M[i][j] == 1)
    {
        M[i][j] = 0;
        DFS(M, i + 1, j, ROW, COL);     //right side traversal
        DFS(M, i - 1, j, ROW, COL);     //left side traversal
        DFS(M, i, j + 1, ROW, COL);     //upward side traversal
        DFS(M, i, j - 1, ROW, COL);     //downward side traversal
        DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
        DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
        DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
        DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
    }
}

int countIslands(vector<vector<int>> &M)
{
    int ROW = M.size();
    int COL = M[0].size();
    int count = 0;
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            if (M[i][j] == 1)
            {
                M[i][j] = 0;
                count++;
                DFS(M, i + 1, j, ROW, COL);     //right side traversal
                DFS(M, i - 1, j, ROW, COL);     //left side traversal
                DFS(M, i, j + 1, ROW, COL);     //upward side traversal
                DFS(M, i, j - 1, ROW, COL);     //downward side traversal
                DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
                DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
                DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
                DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
            }
        }
    }
    return count;
}

// Driver Code
int main()
{
    vector<vector<int>> M = {{1, 1, 0, 0, 0},
                             {0, 1, 0, 0, 1},
                             {1, 0, 0, 1, 1},
                             {0, 0, 0, 0, 0},
                             {1, 0, 1, 0, 1}};

    cout << "Number of islands is: " << countIslands(M);
    return 0;
}

// This code is contributed by ajaymakvana.

Java

// Java Program to count islands in boolean 2D matrix
import java.util.*;
public class Main
{
    // 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 i, int j, int ROW, int COL)
    {

        // Base condition
        // if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL-  or if M[i][j] != 1 then we will simply return
        if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
        {
            return;
        }

        if (M[i][j] == 1)
        {
            M[i][j] = 0;
            DFS(M, i + 1, j, ROW, COL);     //right side traversal
            DFS(M, i - 1, j, ROW, COL);     //left side traversal
            DFS(M, i, j + 1, ROW, COL);     //upward side traversal
            DFS(M, i, j - 1, ROW, COL);     //downward side traversal
            DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
            DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
            DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
            DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
        }
    }

    static int countIslands(int[][] M)
    {
        int ROW = M.length;
        int COL = M[0].length;
        int count = 0;
        for (int i = 0; i < ROW; i++)
        {
            for (int j = 0; j < COL; j++)
            {
                if (M[i][j] == 1)
                {
                    M[i][j] = 0;
                    count++;
                    DFS(M, i + 1, j, ROW, COL);     //right side traversal
                    DFS(M, i - 1, j, ROW, COL);     //left side traversal
                    DFS(M, i, j + 1, ROW, COL);     //upward side traversal
                    DFS(M, i, j - 1, ROW, COL);     //downward side traversal
                    DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
                    DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
                    DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
                    DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
                }
            }
        }
        return count;
    }

  // Driver code
    public static void main(String[] args) {
        int[][] M = {{1, 1, 0, 0, 0},
                 {0, 1, 0, 0, 1},
                 {1, 0, 0, 1, 1},
                 {0, 0, 0, 0, 0},
                 {1, 0, 1, 0, 1}};

        System.out.print("Number of islands is: " + countIslands(M));
    }
}

// This code is contributed by suresh07.

Python3

# Program to count islands in boolean 2D matrix
class Graph:

    def __init__(self, row, col, graph):
        self.ROW = row
        self.COL = col
        self.graph = graph

    # A utility function to do DFS for a 2D
    # boolean matrix. It only considers
    # the 8 neighbours as adjacent vertices
    def DFS(self, i, j):
        if i < 0 or i >= len(self.graph) or j < 0 or j >= len(self.graph[0]) or self.graph[i][j] != 1:
            return

        # mark it as visited
        self.graph[i][j] = -1

        # Recur for 8 neighbours
        self.DFS(i - 1, j - 1)
        self.DFS(i - 1, j)
        self.DFS(i - 1, j + 1)
        self.DFS(i, j - 1)
        self.DFS(i, j + 1)
        self.DFS(i + 1, j - 1)
        self.DFS(i + 1, j)
        self.DFS(i + 1, j + 1)

    # The main function that returns
    # count of islands in a given boolean
    # 2D matrix
    def countIslands(self):
        # Initialize count as 0 and traverse
        # through the all cells of
        # given matrix
        count = 0
        for i in range(self.ROW):
            for j in range(self.COL):
                # If a cell with value 1 is not visited yet,
                # then new island found
                if self.graph[i][j] == 1:
                    # Visit all cells in this island
                    # and increment island count
                    self.DFS(i, j)
                    count += 1

        return count


graph = [
    [1, 1, 0, 0, 0],
    [0, 1, 0, 0, 1],
    [1, 0, 0, 1, 1],
    [0, 0, 0, 0, 0],
    [1, 0, 1, 0, 1]
]


row = len(graph)
col = len(graph[0])

g = Graph(row, col, graph)

print("Number of islands is:", g.countIslands())

# This code is contributed by Shivam Shrey

C#

// C# Program to count islands in boolean 2D matrix
using System;
using System.Collections.Generic;
class GFG {

    // 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 i, int j, int ROW, int COL)
    {

        // Base condition
        // if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL-  or if M[i][j] != 1 then we will simply return
        if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i,j] != 1)
        {
            return;
        }

        if (M[i,j] == 1)
        {
            M[i,j] = 0;
            DFS(M, i + 1, j, ROW, COL);     //right side traversal
            DFS(M, i - 1, j, ROW, COL);     //left side traversal
            DFS(M, i, j + 1, ROW, COL);     //upward side traversal
            DFS(M, i, j - 1, ROW, COL);     //downward side traversal
            DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
            DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
            DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
            DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
        }
    }

    static int countIslands(int[,] M)
    {
        int ROW = M.GetLength(0);
        int COL = M.GetLength(1);
        int count = 0;
        for (int i = 0; i < ROW; i++)
        {
            for (int j = 0; j < COL; j++)
            {
                if (M[i,j] == 1)
                {
                    M[i,j] = 0;
                    count++;
                    DFS(M, i + 1, j, ROW, COL);     //right side traversal
                    DFS(M, i - 1, j, ROW, COL);     //left side traversal
                    DFS(M, i, j + 1, ROW, COL);     //upward side traversal
                    DFS(M, i, j - 1, ROW, COL);     //downward side traversal
                    DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
                    DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
                    DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
                    DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
                }
            }
        }
        return count;
    }

  // Driver code
  static void Main() {
    int[,] M = {{1, 1, 0, 0, 0},
                 {0, 1, 0, 0, 1},
                 {1, 0, 0, 1, 1},
                 {0, 0, 0, 0, 0},
                 {1, 0, 1, 0, 1}};

    Console.Write("Number of islands is: " + countIslands(M));
  }
}

// This code is contributed by decode2207.

Javascript

<script>
    // Javascript Program to count islands in boolean 2D matrix

    // A utility function to do DFS for a 2D
    //  boolean matrix. It only considers
    // the 8 neighbours as adjacent vertices
    function DFS(M, i, j, ROW, COL)
    {
        // Base condition
        // if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL-  or if M[i][j] != 1 then we will simply return
        if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
        {
            return;
        }

        if (M[i][j] == 1)
        {
            M[i][j] = 0;
            DFS(M, i + 1, j, ROW, COL);     //right side traversal
            DFS(M, i - 1, j, ROW, COL);     //left side traversal
            DFS(M, i, j + 1, ROW, COL);     //upward side traversal
            DFS(M, i, j - 1, ROW, COL);     //downward side traversal
            DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
            DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
            DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
            DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
        }
    }

    function countIslands(M)
    {
        let ROW = M.length;
        let COL = M[0].length;
        let count = 0;
        for (let i = 0; i < ROW; i++)
        {
            for (let j = 0; j < COL; j++)
            {
                if (M[i][j] == 1)
                {
                    M[i][j] = 0;
                    count++;
                    DFS(M, i + 1, j, ROW, COL);     //right side traversal
                    DFS(M, i - 1, j, ROW, COL);     //left side traversal
                    DFS(M, i, j + 1, ROW, COL);     //upward side traversal
                    DFS(M, i, j - 1, ROW, COL);     //downward side traversal
                    DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
                    DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
                    DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
                    DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
                }
            }
        }
        return count;
    }

    let M = [[1, 1, 0, 0, 0],
             [0, 1, 0, 0, 1],
             [1, 0, 0, 1, 1],
             [0, 0, 0, 0, 0],
             [1, 0, 1, 0, 1]];

    document.write("Number of islands is: " + countIslands(M));

    // This code is contributed by divyesh072019.
</script>

Output

Number of islands is: 5

Time complexity: O(ROW x COL)
Auxiliary Space: O(1), as we are not using any extra space.

Find the number of Islands | Set 2 (Using Disjoint Set)

Islands in a graph using BFS