<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 {
private:
    class UnionFind {
        private:
            unordered_map<int, int> leader;
        public:
            UnionFind() = default;

            int find(int x) {
                if (x == leader[x]) return x;
                int parent = find(leader[x]);
                leader[x] = parent;
                return parent;
            }

            void add(int x) {
                leader[x] = x;
            }

            void add(int x, int y) {
                if (leader.find(x) == leader.end()) leader[x] = x;
                if (leader.find(y) == leader.end()) leader[y] = y;
                leader[find(x)] = find(y);
            }

            int numGroups() const {
                int ans = 0;
                for (auto &[x, parent] : leader) {
                    if (x == parent) ans++;
                }
                return ans;
            }

            ~UnionFind() = default;
    };

public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        UnionFind myUnion;
        int n = isConnected.size();
        int m = isConnected.size();

        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                if (isConnected[i][j]) {
                    myUnion.add(i, j);
                }
            }
        }

        return myUnion.numGroups();
    }
};
19400