## 1. Problem

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.

**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]

## 2. Solution

I solved this problem like this.

I sovled this problem using dfs. If dfs method called, plus 1 on provinces.

```
class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
provinces = 0
visited = [0] * len(isConnected)
n = len(isConnected)
def dfs(idx):
if visited[idx]:
return
visited[idx] = 1
for i in range(n):
if isConnected[idx][i]:
dfs(i)
for i in range(n):
if not visited[i]:
dfs(i)
provinces += 1
return provinces
```