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Islands in a graph using BFS
  • Difficulty Level : Medium
  • Last Updated : 16 Mar, 2021
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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

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}

This is a variation of the standard problem: 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.
We have discussed a DFS solution for islands is already discussed. This problem can also solved by applying BFS() on each component. In each BFS() call, a component or a sub-graph is visited. We will call BFS on the next un-visited component. The number of calls to BFS() gives the number of connected components. BFS can also be used.

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

C++




// A BFS based solution to count number of
// islands in a graph.
#include <bits/stdc++.h>
using namespace std;
 
// R x C matrix
#define R 5
#define C 5
 
// A function to check if a given cell
// (u, v) can be included in DFS
bool isSafe(int mat[R][C], int i, int j,
            bool vis[R][C])
{
    return (i >= 0) && (i < R) &&
           (j >= 0) && (j < C) &&
           (mat[i][j] && !vis[i][j]);
}
 
void BFS(int mat[R][C], bool vis[R][C],
         int si, int sj)
{
 
    // These arrays are used to get row and
    // column numbers of 8 neighbours of
    // a given cell
    int row[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
    int col[] = { -1, 0, 1, -1, 1, -1, 0, 1 };
 
    // Simple BFS first step, we enqueue
    // source and mark it as visited
    queue<pair<int, int> > q;
    q.push(make_pair(si, sj));
    vis[si][sj] = true;
 
    // Next step of BFS. We take out
    // items one by one from queue and
    // enqueue their univisited adjacent
    while (!q.empty()) {
 
        int i = q.front().first;
        int j = q.front().second;
        q.pop();
 
        // Go through all 8 adjacent
        for (int k = 0; k < 8; k++) {
            if (isSafe(mat, i + row[k],
                       j + col[k], vis)) {
             vis[i + row[k]][j + col[k]] = true;
             q.push(make_pair(i + row[k], j + col[k]));
            }
        }
    }
}
 
// This function returns number islands (connected
// components) in a graph. It simply works as
// BFS for disconnected graph and returns count
// of BFS calls.
int countIslands(int mat[R][C])
{
    // Mark all cells as not visited
    bool vis[R][C];
    memset(vis, 0, sizeof(vis));
 
    // Call BFS for every unvisited vertex
    // Whenever we see an univisted vertex,
    // we increment res (number of islands)
    // also.
    int res = 0;
    for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++) {
            if (mat[i][j] && !vis[i][j]) {
                BFS(mat, vis, i, j);
                res++;
            }
        }
    }
 
    return res;
}
 
// main function
int main()
{
    int mat[][C] = { { 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 << countIslands(mat);
 
    return 0;
}

Java




// A BFS based solution to count number of
// islands in a graph.
import java.util.*;
 
class GFG
{
 
// R x C matrix
static final int R = 5;
static final int C = 5 ;
static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// A function to check if a given cell
// (u, v) can be included in DFS
static boolean isSafe(int mat[][], int i, int j,
                       boolean vis[][])
{
    return (i >= 0) && (i < R) &&
        (j >= 0) && (j < C) &&
        (mat[i][j]==1 && !vis[i][j]);
}
 
static void BFS(int mat[][], boolean vis[][],
                int si, int sj)
{
 
    // These arrays are used to get row and
    // column numbers of 8 neighbours of
    // a given cell
    int row[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
    int col[] = { -1, 0, 1, -1, 1, -1, 0, 1 };
 
    // Simple BFS first step, we enqueue
    // source and mark it as visited
    Queue<pair> q = new LinkedList<pair>();
    q.add(new pair(si, sj));
    vis[si][sj] = true;
 
    // Next step of BFS. We take out
    // items one by one from queue and
    // enqueue their univisited adjacent
    while (!q.isEmpty())
    {
 
        int i = q.peek().first;
        int j = q.peek().second;
        q.remove();
 
        // Go through all 8 adjacent
        for (int k = 0; k < 8; k++)
        {
            if (isSafe(mat, i + row[k],
                    j + col[k], vis))
            {
                vis[i + row[k]][j + col[k]] = true;
                q.add(new pair(i + row[k], j + col[k]));
            }
        }
    }
}
 
// This function returns number islands (connected
// components) in a graph. It simply works as
// BFS for disconnected graph and returns count
// of BFS calls.
static int countIslands(int mat[][])
{
    // Mark all cells as not visited
    boolean [][]vis = new boolean[R][C];
 
    // Call BFS for every unvisited vertex
    // Whenever we see an univisted vertex,
    // we increment res (number of islands)
    // also.
    int res = 0;
    for (int i = 0; i < R; i++)
    {
        for (int j = 0; j < C; j++)
        {
            if (mat[i][j]==1 && !vis[i][j])
            {
                BFS(mat, vis, i, j);
                res++;
            }
        }
    }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int 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 } };
 
    System.out.print(countIslands(mat));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# A BFS based solution to count number of
# islands in a graph.
from collections import deque
 
# A function to check if a given cell
# (u, v) can be included in DFS
def isSafe(mat, i, j, vis):
     
    return ((i >= 0) and (i < 5) and 
            (j >= 0) and (j < 5) and
          (mat[i][j] and (not vis[i][j])))
 
def BFS(mat, vis, si, sj):
     
    # These arrays are used to get row and
    # column numbers of 8 neighbours of
    # a given cell
    row = [-1, -1, -1, 0, 0, 1, 1, 1]
    col = [-1, 0, 1, -1, 1, -1, 0, 1]
 
    # Simple BFS first step, we enqueue
    # source and mark it as visited
    q = deque()
    q.append([si, sj])
    vis[si][sj] = True
 
    # Next step of BFS. We take out
    # items one by one from queue and
    # enqueue their univisited adjacent
    while (len(q) > 0):
        temp = q.popleft()
 
        i = temp[0]
        j = temp[1]
 
        # Go through all 8 adjacent
        for k in range(8):
            if (isSafe(mat, i + row[k], j + col[k], vis)):
                vis[i + row[k]][j + col[k]] = True
                q.append([i + row[k], j + col[k]])
 
# This function returns number islands (connected
# components) in a graph. It simply works as
# BFS for disconnected graph and returns count
# of BFS calls.
def countIslands(mat):
      
    # Mark all cells as not visited
    vis = [[False for i in range(5)]
                  for i in range(5)]
    # memset(vis, 0, sizeof(vis));
 
    # 5all BFS for every unvisited vertex
    # Whenever we see an univisted vertex,
    # we increment res (number of islands)
    # also.
    res = 0
 
    for i in range(5):
        for j in range(5):
            if (mat[i][j] and not vis[i][j]):
                BFS(mat, vis, i, j)
                res += 1
 
    return res
 
# Driver code
if __name__ == '__main__':
     
    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 ]]
 
    print (countIslands(mat))
 
# This code is contributed by mohit kumar 29

C#




// A BFS based solution to count number of
// islands in a graph.
using System;
using System.Collections.Generic;
 
class GFG
{
 
// R x C matrix
static readonly int R = 5;
static readonly int C = 5 ;
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// A function to check if a given cell
// (u, v) can be included in DFS
static bool isSafe(int [,]mat, int i, int j,
                    bool [,]vis)
{
    return (i >= 0) && (i < R) &&
        (j >= 0) && (j < C) &&
        (mat[i, j]==1 && !vis[i, j]);
}
 
static void BFS(int [,]mat, bool [,]vis,
                int si, int sj)
{
 
    // These arrays are used to get row and
    // column numbers of 8 neighbours of
    // a given cell
    int []row = { -1, -1, -1, 0, 0, 1, 1, 1 };
    int []col = { -1, 0, 1, -1, 1, -1, 0, 1 };
 
    // Simple BFS first step, we enqueue
    // source and mark it as visited
    List<pair> q = new List<pair>();
    q.Add(new pair(si, sj));
    vis[si, sj] = true;
 
    // Next step of BFS. We take out
    // items one by one from queue and
    // enqueue their univisited adjacent
    while (q.Count != 0)
    {
        int i = q[0].first;
        int j = q[0].second;
        q.RemoveAt(0);
 
        // Go through all 8 adjacent
        for (int k = 0; k < 8; k++)
        {
            if (isSafe(mat, i + row[k],
                    j + col[k], vis))
            {
                vis[i + row[k], j + col[k]] = true;
                q.Add(new pair(i + row[k], j + col[k]));
            }
        }
    }
}
 
// This function returns number islands (connected
// components) in a graph. It simply works as
// BFS for disconnected graph and returns count
// of BFS calls.
static int countIslands(int [,]mat)
{
    // Mark all cells as not visited
    bool [,]vis = new bool[R, C];
 
    // Call BFS for every unvisited vertex
    // Whenever we see an univisted vertex,
    // we increment res (number of islands)
    // also.
    int res = 0;
    for (int i = 0; i < R; i++)
    {
        for (int j = 0; j < C; j++)
        {
            if (mat[i, j]==1 && !vis[i, j])
            {
                BFS(mat, vis, i, j);
                res++;
            }
        }
    }
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int [,]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 } };
 
    Console.Write(countIslands(mat));
}
}
 
// This code is contributed by 29AjayKumar
Output: 
5

 

Time Complexity : O(V + E) where V is number of vertices and E is number of edges. Note that the given solution is simply works as BFS for disconnected graph.
 

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