<p>You are given an integer <code>n</code>. There is an <strong>undirected</strong> graph with <code>n</code> nodes, numbered from <code>0</code> to <code>n - 1</code>. You are given a 2D integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists an <strong>undirected</strong> edge connecting nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>Return <em>the <strong>number of pairs</strong> of different nodes that are <strong>unreachable</strong> from each other</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/05/tc-3.png" style="width: 267px; height: 169px;" />
<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[0,2],[1,2]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no pairs of nodes that are unreachable from each other. Therefore, we return 0.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/05/tc-2.png" style="width: 295px; height: 269px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
<strong>Output:</strong> 14
<strong>Explanation:</strong> There are 14 pairs of nodes that are unreachable from each other:
[[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]].
Therefore, we return 14.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        pass






















class UnionFind:
    def __init__(self, size):
        self.root = [i for i in range(size)]
        self.rank = [0] * size

    def find(self, x):
        if self.root[x] == 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.rank[rootx] > self.rank[rooty]:
                self.root[rooty] = rootx
            elif self.rank[rootx] < self.rank[rooty]:
                self.root[rootx] = rooty
            else:
                self.root[rootx] = rooty
                self.rank[rooty] += 1
        

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(size=n)

        for edge in edges:
            uf.union(edge[0], edge[1])
        
        component_size = collections.Counter()
        for i in range(n):
            component_size[uf.find(i)] += 1
        
        total_pairs = 0
        remaining_nodes = n
        for i in component_size:
            size = component_size[i]
            total_pairs += size * (remaining_nodes - size)
            remaining_nodes -= size
        return total_pairs
'''
        n_cnts = []
        for key in cnt:
            n_cnts.append(cnt[key])

        n_disc = 0
        for i in range(len(n_cnts)):
            for j in range(i+1, len(n_cnts)):
                n_disc += n_cnts[i] * n_cnts[j]
        
        return n_disc
'''




















class UnionFind:
    def __init__(self, size):
        self.root = [i for i in range(size)]

    def find(self, x):
        if self.root[x] == 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:
            # no merge occurs
            return False
        else:
            # merge occurs
            self.root[rootx] = rooty
            return True

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        n_groups = n
        for e1, e2 in edges:
            if uf.union(e1, e2):
                n_groups -= 1
        group_counter = collections.Counter()
        for node in range(n):
            group_counter[uf.find(node)] += 1

        n_nodes = n
        ans = 0
        print(group_counter, n)
        for key in group_counter:
            n_nodes -= group_counter[key]
            ans += (group_counter[key] * n_nodes)
        return ans






















class UnionFind:
    def __init__(self, n):
        self.roots = [i for i in range(n)]
        self.ranks = [0] * n

    def find(self, x):
        if self.roots[x] == x:
            return x
        self.roots[x] = self.find(self.roots[x])
        return self.roots[x]
    
    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx == rooty:
            return 
        
        if self.ranks[rootx] == self.ranks[rooty]:
            self.ranks[rootx] += 1
            self.roots[rooty] = rootx
        elif self.ranks[rootx] > self.ranks[rooty]:
            self.roots[rooty] = rootx
        else:
            self.roots[rootx] = rooty
        return

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for p1, p2 in edges:
            uf.union(p1, p2)
        
        cnt = collections.defaultdict(int)
        for i in range(n):
            root = uf.find(i)
            cnt[root] += 1
        
        ans = 0
        rem = n
        print(cnt)
        for key in cnt:
            rem -= cnt[key]
            ans += (cnt[key] * rem)
        return ans





















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]
    
    def find(self, x):
        if self.root[x] == 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:
            self.root[rootx] = rooty


class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for a, b in edges:
            uf.union(a, b)
        
        counter_roots = collections.Counter()
        for i in range(n):
            root = uf.find(i)
            counter_roots[root] += 1
        
        ans = 0
        rem = n
        for root in counter_roots:
            amount = counter_roots[root]
            rem -= amount
            ans += (amount * rem)
        return ans

















class UnionFind:
    def __init__(self, n):
        self.roots = [i for i in range(n)]

    def find(self, x):
        if x == self.roots[x]:
            return x
        self.roots[x] = self.find(self.roots[x])
        return self.roots[x]

    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            self.roots[rootx] = rooty

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for a, b in edges:
            uf.union(a, b)

        cnt = collections.Counter()
        for i in range(n):
            cnt[uf.find(i)] += 1
        
        rem = n
        ans = 0
        for i in cnt:
            rem -= cnt[i]
            ans += cnt[i] * (rem)
        return ans




















class UnionFind:
    def __init__(self, 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:
            return
        else:
            self.root[rootx] = rooty
            return

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for a, b in edges:
            uf.union(a, b)
        
        n_groups = collections.Counter()
        for node in range(n):
            n_groups[uf.find(node)] += 1
        
        ans = 0
        total_node = n
        for key in n_groups:
            ans += n_groups[key] * (total_node - n_groups[key])
        return ans // 2

















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]
    
    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:
            return
        self.root[rootx] = rooty

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for a, b in edges:
            uf.union(a, b)
        
        n_groups = collections.Counter()
        for i in range(n):
            n_groups[uf.find(i)] += 1
        
        ans = 0
        for key in n_groups.keys():
            ans += (n_groups[key] * (n - n_groups[key]))
        return ans // 2
78400