Number of Provinces

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

Solution

BFS approach:

Recursively traverse every node and add every neighbor that’s 1 into the stack, mark traversed node visited; Keep a counter i for number of provinces;

After the stack is empty, increment i and find the next unvisited 1 in the matrix and repeat that traverse.

Exit the loop and return i as the number of provinces

class Solution:
    # helper method for adding every connected neightbor to the stack
    @staticmethod
    def bfs(m, visited, i):
        stack = [i]
        while(len(stack) != 0):
            n = stack.pop(0) # pop from the stack
            visited[n] = 1
            # visit every unvisited neighbor to check if it is connected
            # check if the nth column of every row is 1
            for i in range(len(m)):
                if not visited[i]:
                    if m[n][i]:
                        stack.append(i)
        
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        visited = [0] * len(isConnected)
        count = 0
        for i in range(len(isConnected)):
            if not visited[i]:
                count += 1
                self.bfs(isConnected, visited, i)
        return count