"""

# https://xiaoguan.gitbooks.io/leetcode/content/LeetCode/547-friend-circles-medium.html

547. Friend Circles (Medium)
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.
Note:

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

"""

# V0 

# V1 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : disjoint- set 
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        N = len(M)
        f = range(N)

        def find(x):
            while f[x] != x: x = f[x]
            return x

        for x in range(N):
            for y in range(x + 1, N):
                if M[x][y]: f[find(x)] = find(y)
        return sum(f[x] == x for x in range(N))

# V1' 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : DFS  (stack)
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        cnt, N = 0, len(M)
        vset = set()
        def dfs(n):
            for x in range(N):
                if M[n][x] and x not in vset:
                    vset.add(x)
                    dfs(x)
        for x in range(N):
            if x not in vset:
                cnt += 1
                dfs(x)
        return cnt

# V1'' 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : BFS (queue)
lass Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        cnt, N = 0, len(M)
        vset = set()
        def bfs(n):
            q = [n]
            while q:
                n = q.pop(0)
                for x in range(N):
                    if M[n][x] and x not in vset:
                        vset.add(x)
                        q.append(x)
        for x in range(N):
            if x not in vset:
                cnt += 1
                bfs(x)
        return cnt

# V1'''
# https://www.jiuzhang.com/solution/friend-circles/#tag-highlight-lang-python
# IDEA : UNION FIND 
class Solution:
    """
    @param M: a matrix
    @return: the total number of friend circles among all the students
    """
    def BFS(self, student, M):
        queue = []
        queue.append(student)
        while len(queue) :
            size = len(queue)
            for i in range(0, size) :    # MAKE SURE THE DISTANCE BETWEEN START, END POINT ARE THE SAME DURING EVERY SRARCH 
                j = queue[0]
                del queue[0]
                M[j][j] = 2
                for k in range(0, len(M[0])):   # GO THROUGH ALL FRIENDSHIPS 
                    if M[j][k] == 1 and M[k][k] == 1:   # IF M[k][k]==1 -> k is not searched yet, KEEP SEARCHING  
                        queue.append(k)
    def findCircleNum(self, M):
        # Write your code here
        count = 0
        for i in range(0, len(M)):
            if M[i][i] == 1 :   # IF CURRENT diagonal =1 -> THIS PERSON IS IN THE NEW CIRCLE 
                count += 1  # COUNT + 1 
                self.BFS(i, M) # START SEARCH 
        return count

# V2 
# Time:  O(n^2)
# Space: O(n)
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """

        class UnionFind(object):
            def __init__(self, n):
                self.set = range(n)
                self.count = n

            def find_set(self, x):
               if self.set[x] != x:
                   self.set[x] = self.find_set(self.set[x])  # path compression.
               return self.set[x]

            def union_set(self, x, y):
                x_root, y_root = map(self.find_set, (x, y))
                if x_root != y_root:
                    self.set[min(x_root, y_root)] = max(x_root, y_root)
                    self.count -= 1

        circles = UnionFind(len(M))
        for i in range(len(M)):
            for j in range(len(M)):
                if M[i][j] and i != j:
                    circles.union_set(i, j)
        return circles.count
"""

# https://xiaoguan.gitbooks.io/leetcode/content/LeetCode/547-friend-circles-medium.html

547. Friend Circles (Medium)
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.
Note:

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

"""

# V0 

# V1 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : disjoint- set 
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        N = len(M)
        f = range(N)

        def find(x):
            while f[x] != x: x = f[x]
            return x

        for x in range(N):
            for y in range(x + 1, N):
                if M[x][y]: f[find(x)] = find(y)
        return sum(f[x] == x for x in range(N))

# V1' 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : DFS  (stack)
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        cnt, N = 0, len(M)
        vset = set()
        def dfs(n):
            for x in range(N):
                if M[n][x] and x not in vset:
                    vset.add(x)
                    dfs(x)
        for x in range(N):
            if x not in vset:
                cnt += 1
                dfs(x)
        return cnt

# V1'' 
# https://blog.csdn.net/XX_123_1_RJ/article/details/82656277
# http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/
# IDEA : BFS (queue)
lass Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """
        cnt, N = 0, len(M)
        vset = set()
        def bfs(n):
            q = [n]
            while q:
                n = q.pop(0)
                for x in range(N):
                    if M[n][x] and x not in vset:
                        vset.add(x)
                        q.append(x)
        for x in range(N):
            if x not in vset:
                cnt += 1
                bfs(x)
        return cnt

# V1'''
# https://www.jiuzhang.com/solution/friend-circles/#tag-highlight-lang-python
# IDEA : UNION FIND 
class Solution:
    """
    @param M: a matrix
    @return: the total number of friend circles among all the students
    """
    def BFS(self, student, M):
        queue = []
        queue.append(student)
        while len(queue) :
            size = len(queue)
            for i in range(0, size) :    # MAKE SURE THE DISTANCE BETWEEN START, END POINT ARE THE SAME DURING EVERY SRARCH 
                j = queue[0]
                del queue[0]
                M[j][j] = 2
                for k in range(0, len(M[0])):   # GO THROUGH ALL FRIENDSHIPS 
                    if M[j][k] == 1 and M[k][k] == 1:   # IF M[k][k]==1 -> k is not searched yet, KEEP SEARCHING  
                        queue.append(k)
    def findCircleNum(self, M):
        # Write your code here
        count = 0
        for i in range(0, len(M)):
            if M[i][i] == 1 :   # IF CURRENT diagonal =1 -> THIS PERSON IS IN THE NEW CIRCLE 
                count += 1  # COUNT + 1 
                self.BFS(i, M) # START SEARCH 
        return count

# V2 
# Time:  O(n^2)
# Space: O(n)
class Solution(object):
    def findCircleNum(self, M):
        """
        :type M: List[List[int]]
        :rtype: int
        """

        class UnionFind(object):
            def __init__(self, n):
                self.set = range(n)
                self.count = n

            def find_set(self, x):
               if self.set[x] != x:
                   self.set[x] = self.find_set(self.set[x])  # path compression.
               return self.set[x]

            def union_set(self, x, y):
                x_root, y_root = map(self.find_set, (x, y))
                if x_root != y_root:
                    self.set[min(x_root, y_root)] = max(x_root, y_root)
                    self.count -= 1

        circles = UnionFind(len(M))
        for i in range(len(M)):
            for j in range(len(M)):
                if M[i][j] and i != j:
                    circles.union_set(i, j)
        return circles.count
# https://xiaoguan.gitbooks.io/leetcode/content/LeetCode/547-friend-circles-medium.html  547. Friend Circles (Medium) 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. Note:  N is in range [1,200]. M[i][i] = 1 for all students. If M[i][j] = 1, then M[j][i] = 1. # V0 # V1 # https://blog.csdn.net/XX_123_1_RJ/article/details/82656277 # http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/ # IDEA : disjoint- set :type M: List[List[int]]         :rtype: int # V1' # https://blog.csdn.net/XX_123_1_RJ/article/details/82656277 # http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/ # IDEA : DFS  (stack) :type M: List[List[int]]         :rtype: int # V1'' # https://blog.csdn.net/XX_123_1_RJ/article/details/82656277 # http://bookshadow.com/weblog/2017/04/03/leetcode-friend-circles/ # IDEA : BFS (queue) :type M: List[List[int]]         :rtype: int # V1''' # https://www.jiuzhang.com/solution/friend-circles/#tag-highlight-lang-python # IDEA : UNION FIND @param M: a matrix     @return: the total number of friend circles among all the students # MAKE SURE THE DISTANCE BETWEEN START, END POINT ARE THE SAME DURING EVERY SRARCH # GO THROUGH ALL FRIENDSHIPS # IF M[k][k]==1 -> k is not searched yet, KEEP SEARCHING # Write your code here # IF CURRENT diagonal =1 -> THIS PERSON IS IN THE NEW CIRCLE # COUNT + 1 # START SEARCH # V2 # Time:  O(n^2) # Space: O(n) :type M: List[List[int]]         :rtype: int # path compression.