<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 {
public:
    void perform_bfs(vector<vector<int>> &adj_list, int node, vector<int> &visited) {

        queue<int> q;
        q.push(node);
        visited[node] = 1;

        while(!q.empty()) {
            int curr_node = q.front();
            q.pop();

            for (auto adj_node : adj_list[curr_node]) {
                if (visited[adj_node] == 0)
                    q.push(adj_node);

                visited[adj_node] = 1;
            }
        }
    }
    int findCircleNum(vector<vector<int>>& isConnected) {

        // Here we have an adjacency matrix implementation, Just do a BFS/DFS to get the number of provinces

        // First lets convert this into an adjacency list implementation
        vector<vector<int>> adj_list(isConnected.size(), vector<int>(isConnected[0].size(), 0));
        // adj_list = {0};
        vector<int> visited(isConnected.size(), 0);


        for (int i = 0; i<isConnected.size(); i++) {
            for (int j = 0; j<isConnected[0].size(); j++) {
                if (isConnected[i][j] == 1)
                    adj_list[i].push_back(j);
            }
        }

        int provinces = 0;

        // Once the adjacency_list is created, lets call for BFS
        for (int i = 0; i<isConnected.size(); i++) {
            if (visited[i] == 0) {
                perform_bfs(adj_list, i, visited);
                provinces++;
            }
        }

        return provinces;
        
    }
};
22200