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

<p>A <strong>province</strong> is a group of directly or indirectly connected cities and no other cities outside of the group.</p>

<p>You are given an <code>n x n</code> matrix <code>isConnected</code> where <code>isConnected[i][j] = 1</code> if the <code>i<sup>th</sup></code> city and the <code>j<sup>th</sup></code> city are directly connected, and <code>isConnected[i][j] = 0</code> otherwise.</p>

<p>Return <em>the total number of <strong>provinces</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/24/graph1.jpg" style="width: 222px; height: 142px;" />
<pre>
<strong>Input:</strong> isConnected = [[1,1,0],[1,1,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/24/graph2.jpg" style="width: 222px; height: 142px;" />
<pre>
<strong>Input:</strong> isConnected = [[1,0,0],[0,1,0],[0,0,1]]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>n == isConnected.length</code></li>
	<li><code>n == isConnected[i].length</code></li>
	<li><code>isConnected[i][j]</code> is <code>1</code> or <code>0</code>.</li>
	<li><code>isConnected[i][i] == 1</code></li>
	<li><code>isConnected[i][j] == isConnected[j][i]</code></li>
</ul>

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def dfs(city:int):
            stack = [city]
            while stack:
                node = stack.pop()
                visited[node] = True
                for adj in range(n):
                    if isConnected[node][adj] and not visited[adj]:
                        stack.append(adj)

        n = len(isConnected)
        visited = [False]*n
        provinces = 0
        for city in range(n):
            if visited[city]: continue
            provinces += 1
            dfs(city) 


        return provinces
        # n = len(isConnected)
        # def markProvince(i:int, j:int):
        #     stack = [i,j]
        #     while stack:
        #         city = stack.pop()
        #         for adj in range(n):
        #             if isConnected[city][adj]:
        #                 stack.append(adj)
        #                 isConnected[city][adj] = 0
        #                 isConnected[adj][city] = 0


        # provinces = 0
        # for i in range(n):
        #     for j in range(i,n):
        #         if isConnected[i][j]:
        #             provinces += 1
        #             # make province as seen with -1
        #             markProvince(i,j)
        # return provinces

16800