# Find the number of Islands Using Disjoint Set in C++

C++Server Side ProgrammingProgramming

In this problem, we are given a 2D binary matrix. Our task is to find the number of islands Using DFS.

Island is a ground of 1 or more connected 1’s in the matrix.

Let’s take an example to understand the problem,

Input

bin[][] = {{ 1 0 0 0}
{0 1 0 1}
{0 0 0 0}
{0 0 1 0}}

Output

3


Explanation

Islands are :
bin00 - bin11
bin13
bin32

## Solution Approach

To find the island from a binary matrix using a disjoint set data structure. To find island count, we will traverse the matrix and do union of all adjacent vertices by checking all 8 neighbours, if they are 1 take union of current index with its neighbour. Then do the second traversal of the matrix, and if at any index the value is 1, find its sent. If frequency is 0, increase to 1.

## Example

Program to illustrate the working of our solution,

#include <bits/stdc++.h>
using namespace std;
class DisjointUnionSets{
vector<int> rank, parent;
int n;
public:
DisjointUnionSets(int n){
rank.resize(n);
parent.resize(n);
this->n = n;
makeSet();
}
void makeSet(){
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int x){
if (parent[x] != x){
return find(parent[x]);
}
return x;
}
void Union(int x, int y){
int xRoot = find(x);
int yRoot = find(y);
if (xRoot == yRoot)
return;
if (rank[xRoot] < rank[yRoot])
parent[xRoot] = yRoot;
else if (rank[yRoot] < rank[xRoot])
parent[yRoot] = xRoot;
else {
parent[yRoot] = xRoot;
rank[xRoot] = rank[xRoot] + 1;
}
}
};
int findIslandCount(vector<vector<int>> mat){
int n = mat.size();
int m = mat[0].size();
DisjointUnionSets *dus = new DisjointUnionSets(n * m);
for (int j = 0; j < n; j++){
for (int k = 0; k < m; k++){
if (mat[j][k] == 0)
continue;
if (j + 1 < n && mat[j + 1][k] == 1)
dus->Union(j * (m) + k, (j + 1) * (m) + k);
if (j - 1 >= 0 && mat[j - 1][k] == 1)
dus->Union(j * (m) + k, (j - 1) * (m) + k);
if (k + 1 < m && mat[j][k + 1] == 1)
dus->Union(j * (m) + k, (j) * (m) + k + 1);
if (k - 1 >= 0 && mat[j][k - 1] == 1)
dus->Union(j * (m) + k, (j) * (m) + k - 1);
if (j + 1 < n && k + 1 < m && mat[j + 1][k + 1] == 1)
dus->Union(j * (m) + k, (j + 1) * (m) + k + 1);
if (j + 1 < n && k - 1 >= 0 && mat[j + 1][k - 1] == 1)
dus->Union(j * m + k, (j + 1) * (m) + k - 1);
if (j - 1 >= 0 && k + 1 < m && mat[j - 1][k + 1] == 1)
dus->Union(j * m + k, (j - 1) * m + k + 1);
if (j - 1 >= 0 && k - 1 >= 0 && mat[j - 1][k - 1] == 1)
dus->Union(j * m + k, (j - 1) * m + k - 1);
}
}
int *c = new int[n * m];
int islands = 0;
for (int j = 0; j < n; j++){
for (int k = 0; k < m; k++){
if (mat[j][k] == 1){
int x = dus->find(j * m + k);
if (c[x] == 0){
islands++;
c[x]++;
}
else
c[x]++;
}
}
}
return islands;
}
int main(void){
vector<vector<int>> mat = {
{1, 1, 0, 1, 0},
{0, 1, 0, 1, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 1, 1, 0, 1}
};
cout<<"The number of islands in binary matrix is : "<<findIslandCount(mat);
}

## Output

The number of islands in binary matrix is : 4
Updated on 11-Feb-2022 10:58:15