Problem description:

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:

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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:

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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.

Note:
N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

Solution BFS:

The idea is, when we visited a person, traverse all his friends and other related people. THe following solution use BFS concept. if(!visited[j] && M[tmp][j]) means we only need to check a person when it has not been visited yet, and he is a friend of tmp.

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class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        //BFS
        int n= M.size(), res= 0;
        queue<int> q;
        vector<int> visited(n, 0);
        for(int i= 0; i< n; i++){
            if(visited[i]) continue;
            q.push(i);
            while(!q.empty()){
                int tmp= q.front();
                q.pop();
                visited[tmp]= true;
                for(int j= 0; j< n; j++){
                    //if(visited[j] || !M[tmp][j]) continue;
                    if(visited[j]) continue;
                    if(!visited[j] && M[tmp][j])
                        q.push(j);
                }
            }
            res++;
        }
        return res;
    }
};

time complexity: $O(n^2)$
space complexity: $O(n)$

Solution DFS:

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class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def dfs(i):
            for r in range(n):
                if isConnected[r][i] and visited[r] == False:
                    visited[r] = True
                    dfs(r)
        
        res = 0
        n = len(isConnected)
        visited = [False]*n
        for i in range(n):
            if visited[i] == False:
                res += 1
                visited[i] = True
                dfs(i)
        
        return res
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class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        //BFS
        int n= M.size(), res= 0;
        queue<int> q;
        vector<int> visited(n, 0);
        for(int i= 0; i< n; i++){
            if(visited[i]) continue;
            dfs(M, visited, i);
            res++;
        }
        return res;
    }
    void dfs(vector<vector<int>>& M, vector<int>& visited, int index){
        if(visited[index]) return;
        visited[index]= true;
        for(int i= 0; i< M.size(); i++){
            if(visited[i]) continue;
            if(!visited[i] && M[index][i])
                dfs(M, visited, i);
        }
    }
};

time complexity: $O(n^2)$
space complexity: $O(n)$

Solution Union Find:

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class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        parent = [i for i in range(n)]
        rank = [1]*n # calculate how many node in this group, init is 1 since everyone is own group

        def find(i):
            if parent[i] != i:
                return find(parent[i])
            return parent[i]
        
        def union(i, j):
            rooti, rootj = find(i), find(j)
            isConnected[i][j], isConnected[j][i] = 0, 0
            if rooti == rootj:
                return
            
            rank_big, rank_small = (rooti, rootj) if rank[i] > rank[j] else (rootj, rooti)
            parent[rank_small] = rank_big
            rank[rank_big] += rank[rank_small]
        
        for i in range(n):
            for j in range(n):
                if isConnected[i][j]:
                    union(i, j)
        return len(set(find(i) for i in range(n)))

reference: