547. Friend Circles

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.
Example 1:

Input: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
The 2nd student himself is in a friend circle. So return 2.

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.

https://leetcode.com/problems/friend-circles/solution/


// from leetcode solution 

// dfs

public class Solution {
    public void dfs(int[][] M, int[] visited, int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && visited[j] == 0) {
                visited[j] = 1;
                dfs(M, visited, j);
            }
        }
    }
    public int findCircleNum(int[][] M) {
        int[] visited = new int[M.length];
        int count = 0;
        for (int i = 0; i < M.length; i++) {
            if (visited[i] == 0) {
                dfs(M, visited, i);
                count++;
            }
        }
        return count;
    }
}

// uf 
This is a typical Union Find problem. I abstracted it as a standalone class. Remember the template, you will be able to use it later.

public class Solution {
    class UnionFind {
        private int count = 0;
        private int[] parent, rank;
        
        public UnionFind(int n) {
            count = n;
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }
        
        public int find(int p) {
            while (p != parent[p]) {
                parent[p] = parent[parent[p]];    // path compression by halving
                p = parent[p];
            }
            return p;
        }
        
        public void union(int p, int q) {
            int rootP = find(p);
            int rootQ = find(q);
            if (rootP == rootQ) return;
            if (rank[rootQ] > rank[rootP]) {
                parent[rootP] = rootQ;
            }
            else {
                parent[rootQ] = rootP;
                if (rank[rootP] == rank[rootQ]) {
                    rank[rootP]++;
                }
            }
            count--;
        }
        
        public int count() {
            return count;
        }
    }
    
    public int findCircleNum(int[][] M) {
        int n = M.length;
        UnionFind uf = new UnionFind(n);
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (M[i][j] == 1) uf.union(i, j);
            }
        }
        return uf.count();
    }
}

// bfs 

Equivalent to Finding Connected Components in a Graph

public int findCircleNum(int[][] M) {
    int count = 0;
    for (int i=0; i<M.length; i++)
        if (M[i][i] == 1) { count++; BFS(i, M); }
    return count;
}

public void BFS(int student, int[][] M) {
    Queue<Integer> queue = new LinkedList<>();
    queue.add(student);
    while (queue.size() > 0) {
        int queueSize = queue.size();
        for (int i=0;i<queueSize;i++) {
            int j = queue.poll();
            M[j][j] = 2; // marks as visited
            for (int k=0;k<M[0].length;k++) 
                if (M[j][k] == 1 && M[k][k] == 1) queue.add(k);
        }
    }
}