<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:
        from collections import defaultdict
        
        # Step 1: Create an adjacency list for the graph
        graph = defaultdict(list)
        for n1, n2 in edges:
            graph[n1].append(n2)
            graph[n2].append(n1)
        
        # Step 2: Function to perform DFS and return the size of the connected component
        def dfs(node):
            visited.add(node)
            component_size = 1
            for neighbor in graph[node]:
                if neighbor not in visited:
                    component_size += dfs(neighbor)
            return component_size
        
        visited = set()
        total_pairs = 0
        total_nodes_in_components = 0
        
        # Step 3: Perform DFS for each unvisited node
        for i in range(n):
            if i not in visited:
                # Get the size of the current connected component
                component_size = dfs(i)
                
                # Calculate unreachable pairs for the current component
                unreachable_pairs = component_size * (n - component_size)
                
                # Accumulate the total unreachable pairs
                total_pairs += unreachable_pairs
                
                # Track the number of nodes processed so far
                total_nodes_in_components += component_size
        
        # Step 4: Since we counted each pair twice, return half the total_pairs
        return total_pairs // 2

1687