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 `i`

city and the ^{th}`j`

city are directly connected, and ^{th}`isConnected[i][j] = 0`

otherwise.

Return *the total number of provinces*.

**Example 1:**

Input:isConnected = [[1,1,0],[1,1,0],[0,0,1]]Output:2

**Example 2:**

Input:isConnected = [[1,0,0],[0,1,0],[0,0,1]]Output:3

**Constraints:**

`1 <= n <= 200`

`n == isConnected.length`

`n == isConnected[i].length`

`isConnected[i][j]`

is`1`

or`0`

.`isConnected[i][i] == 1`

`isConnected[i][j] == isConnected[j][i]`

From some source, we can visit every connected node to it with a simple DFS. As is the case with DFS’s, **seen** will keep track of nodes that have been visited.

For every node, we can visit every node connected to it with this DFS, and increment our answer as that represents one friend circle (connected component.)

class Solution: def findCircleNum(self, A): N = len(A) seen = set() def dfs(node): for nei, adj in enumerate(A[node]): if adj and nei not in seen: seen.add(nei) dfs(nei) ans = 0 for i in range(N): if i not in seen: dfs(i) ans += 1 return ans