<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:
    unordered_map<int,bool> visited;
    void bfs(int node, unordered_map<int,list<int>>& adj){
        visited[node] = true;
        queue<int> q;
        q.push(node);
        while(!q.empty()){
            int front = q.front();
            q.pop();
            for(auto neighbours: adj[front]){
                if(!visited[neighbours]){
                    visited[neighbours] = true;
                    q.push(neighbours);
                }
            }
        }
    }
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int count = 0;
        unordered_map<int,list<int>> adj;
        for(int i = 0; i<n; i++){
            for(int j = 0; j<n; j++){
                if(isConnected[i][j] == 1 && i!=j){
                    adj[i].push_back(j);
                    adj[j].push_back(i);
                }
            }
        }
        for(int i = 0; i<n; i++){
            if(!visited[i]){
                bfs(i,adj);
                count++;
            }
        }
    return count;

    }
};
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