<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 UnionFind:
    def __init__(self, size):
        self.size = [1]*(size)
        self.root = [i for i in range(size)]
        
        
    def find(self, x):
        if x == self.root[x]:
            return  x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.size[rootX] > self.size[rootY]:
                self.root[rootY] = rootX
                self.size[rootX] += self.size[rootY]
            else:
                self.root[rootX] = rootY
                self.size[rootY] += self.size[rootX]
            return True
        return False
    def connected(self, x, y):
        return self.find(x) == self.find(y)
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        n = len(isConnected)
        obj = UnionFind(n)
        def isbound(row , col):
            return 0 <= row < n and 0 <= col < n
        visit = []
        res = n
        for i in range(n):
            for j in range(i+1 , n):
                if isConnected[i][j] == 1:
                    res -= obj.union(i , j)
        print(obj.root)
        return res

        
17300