<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 DisjointSet{
public:
    vector<int> parent, size;
    DisjointSet (int n){
        parent.resize(n+1);
        size.resize(n+1);
        for (int i=0; i<n+1; i++){
            size[i] = 1;
            parent[i] = i;
        }
    }

    int findUParent(int node){
        if (node == parent[node]){
            return node;
        }

        return parent[node] = findUParent(parent[node]);
    }

    void makeConnection(int u, int v){
        int up_U = findUParent(u), up_V = findUParent(v);
        if(up_U == up_V)  return;
        else if (size[up_U] >= size[up_V]){
            size[up_U] += size[up_V];
            parent[up_V] = up_U;
        }
        else{
            size[up_V] += size[up_U];
            parent[up_U] = up_V;
        }
    }
};

class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        // creating edges
        vector<vector<int>> edges;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(i!=j && isConnected[i][j]==1){
                    edges.push_back({i, j});
                }
            }
        }

        DisjointSet ds(n);
        for (auto it : edges){
            ds.makeConnection(it[0], it[1]);
        }

        int cntProvinces = 0;
        for (int i=0; i<n; i++){
            if (ds.findUParent(i) == i)  cntProvinces++;
        }
        return cntProvinces;
    }
};
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