<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:
#         # in order to get the number of pairs
#         # we need to group nodes by their connections
#         # the nodes that can reach each other are in the same cluster
#         # then the output is just the combo between different clusters

#         # to find each cluster, we first create the graph, and then traverse the nodes for each cluster using BFS
#         graph = {}
#         for edge in edges:
#             if edge[0] not in graph:
#                 graph[edge[0]] = [edge[1]]
#             else:
#                 graph[edge[0]].append(edge[1])
#             if edge[1] not in graph:
#                 graph[edge[1]] = [edge[0]]
#             else:
#                 graph[edge[1]].append(edge[0])
        
#         traverse = [0] * n
#         clusters = []
#         for i in range(n):
#             # go over each node
#             # create a new cluster if traverse[i] == 0
#             if traverse[i] == 0:
#                 # create new cluster
#                 traverse[i] = 1
#                 clusters.append([i])
#                 # BFS
#                 if i in graph:                    
#                     nodes_curr = graph[i]
#                     while nodes_curr:
#                         nodes_next = []
#                         for node in nodes_curr:
#                             if traverse[node] == 0:
#                                 traverse[node] = 1
#                                 clusters[-1].append(node)
#                                 nodes_next += graph[node]
#                         nodes_curr = nodes_next
#         ans = 0
#         for i in range(len(clusters)):
#             for j in range(i+1, len(clusters)):
#                 ans+=len(clusters[i]) * len(clusters[j])
#         return ans

from collections import defaultdict
class Solution:
    def countPairs(self, n: int, e: List[List[int]]) -> int:
        visit = [False] * n
        edges = defaultdict(list)
        comp = []

        def dfs(curr):
            inside = 1
            for i in edges[curr]:
                if not visit[i]:
                    visit[i] = True
                    inside += dfs(i)
            return inside

        for i in e:
            edges[i[0]].append(i[1])
            edges[i[1]].append(i[0])
        
        for i in range(n):
            if not visit[i]:
                visit[i] = True
                comp.append(dfs(i))
        
        res = 0
        for i in comp:
            res += ((n - i) * i)
            n -= i
        return res


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