Create 547. Friend Circles.md

Author: syhwawa

Leetcode practice/547. Friend Circles.md

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+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.
+
+Constraints:
+
+1 <= N <= 200
+M[i][i] == 1
+M[i][j] == M[j][i]
+
+```Python
+class Solution:
+    def findCircleNum(self, M: List[List[int]]) -> int:
+        N = len(M)
+        visited = [0] *N
+        count = 0
+        queue = []
+        
+        arr = [i for i in range(N)]
+        while(len(arr) > 0):
+		   # Starting a new cycle
+            queue.append(arr.pop(0))  
+            while len(queue) > 0:
+                i = queue.pop(0)
+                if visited[i] == 0:
+                    count +=1
+                    visited[i] = 1
+                    # Check all neighboring nodes, adding to queueu
+                for j, value in enumerate(M[i]):
+                    if value == 1 and visited[j] == 0:
+                        visited[j] = 1
+                        queue.append(j)
+                        arr.remove(j)
+                if i in arr:
+                    arr.remove(i)
+            
+        return count
+
+```