Leetcode 547. Friend Circles (Python)

Related Topic

Depth-First-Search. Tree.

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.

Sample I/O

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.

Note

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

Methodology

This question can be solved by Depth First Search.

We are using the dfs to find the students’ friends and the student’s friends’ friends and so on and store them as 1 friend circle. If other student in this circle then we skip until we find a student not in this circle, this student will start another circle.

Code

def findCircleNum(self, M: List[List[int]]) -> int:
        visited = set()
        count = 0
        def dfs(student, visited):
            for class_mate, is_friend in enumerate(M[student]):
                if class_mate not in visited and is_friend:
                    visited.add(class_mate)
                    dfs(class_mate, visited)
                    
        for student in range(len(M)):
            if student not in visited:
                visited.add(student)
                count+=1
                dfs(student, visited)
        return count

BigO

We traversal all students and all students' classmates so total time complexity is O(n^2)