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

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 graph with only one connected component is called as 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 5 islands

{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}

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 <stdio.h> #include <string.h> #include <stdbool.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 travese 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 travese 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

## Python

# 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 travese # 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

Output:

Number of islands is: 5

Time complexity: O(ROW x COL)

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

Reference:

http://en.wikipedia.org/wiki/Connected_component_%28graph_theory%29

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