Problem

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

OR

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Examples

Example 1:

graph LR
1 --- 2
3
  
Input:
isConnected =[[1,1,0],[1,1,0],[0,0,1]]
Output:
 2

Example 2:

graph TD
1
2
3
  
Input:
isConnected =[[1,0,0],[0,1,0],[0,0,1]]
Output:
 3

Solution

Method 1 - DFS

Code

Java
public class Solution {

	public int findCircleNum(int[][] isConnected) {
		boolean[] visited = new boolean[isConnected.length];
		int count = 0;

		for (int i = 0; i < isConnected.length; i++) {
			if (!visited[i]) {
				dfs(isConnected, visited, i);
				count++;
			}
		}

		return count;
	}

	public void dfs(int[][] isConnected, boolean[] visited, int i) {
		for (int j = 0; j < isConnected.length; j++) {
			if (isConnected[i][j] == 1 && !visited[j]) {
				visited[j] = true;
				dfs(isConnected, visited, j);
			}
		}
	}

}

Complexity

  • Time: O(n^2) for doing dfs as it is adjacency matrix representation
  • Space: O(n) for recursion stack and also visited array.

Method 2 - Union Find Data Structure

This is a typical Union Find problem. Just add all the edges to it, and count the number of components.

Code

Java
public class Solution {

	public int findCircleNum(int[][] isConnected) {
		int n = isConnected.length;
		UnionFind uf = new UnionFind(n);

		for (int i = 0; i < n - 1; i++) {
			for (int j = i + 1; j < n; j++) {
				if (isConnected[i][j] == 1) {
					uf.union(i, j);
				}
			}
		}

		return uf.getNumComponents();
	}

	static class UnionFind {

		private int[] parent;

		private int[] rank;

		private int numComponents;

		public UnionFind(int n) {
			this.numComponents = n;
			makeSet();
		}

		private void makeSet() {
			rank = new int[numComponents];
			parent = new int[numComponents];

			for (int i = 0; i < parent.length; i++) {
				parent[i] = i;
				rank[i] = 0;
			}
		}

		public boolean union(int x, int y) {
			int root1 = find(x);
			int root2 = find(y);

			if (root1 == root2) {
				return false;
			}

			if (rank[root1] > rank[root2]) {
				parent[root2] = root1;
			} else {
				parent[root1] = root2;
			}

			if (rank[root1] == rank[root2]) {
				rank[root2] += 1;
			}

			numComponents--;
			return true;
		}

		public int find(int x) {
			while (parent[x] != x) {
				x = parent[x];
			}

			return x;
		}
		// count
		public int getNumComponents() {
			return numComponents;
		}

		public boolean isFullyConnected() {
			return numComponents == 1;
		}
	}
}

Complexity

  • ⏰ Time complexity: O(n^2 * α(n))
    • There are two nested loops which iterate over the upper triangular matrix excluding the diagonal.
      • Outer loop (i): runs from 0 to n-2 -> O(n)
      • Inner loop (j): runs from i+1 to n-1 -> O(n-i-1)
    • The union and find operations on average take O(α(n)), where α is the inverse Ackermann function, which is very close to O(1) for practical values of n.
  • 🧺 Space complexity: O(n) for storing parent and rank array.