<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:
        '''
        The code blocks mentioned w/i bfs belongs to the bfs algo.
        The code blocks mentioned w/i dfs belongs to the dfs algo.
        The remaining code is common for both the algos
        '''
        # dfs start
        def dfs(src, graph):
            nonlocal size
            visited[src] = 1
            size += 1
            for neighbor in graph[src]:
                if visited[neighbor] == 0:
                    dfs(neighbor, graph)
        # dfs end
        
        graph = collections.defaultdict(list)
        rem_nodes = N

        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)

        ans = 0
        visited = [0]*N
        q = deque()

        
        for i in range(N):
            if visited[i] == 0:
                size = 0
                # dfs start
                dfs(i, graph)
                ans += size*(rem_nodes-size)
                rem_nodes -= size
                # dfs end

                # bfs start

                # visited[i] = 1

                # q.append(i)

                # # start bfs
                # while q:
                #     node = q.popleft()
                #     size += 1

                #     for neighbor in graph[node]:
                #         if visited[neighbor] == 0:
                #             visited[neighbor] = 1
                #             q.append(neighbor)
                
                # ans += size * (rem_nodes - size)
                # rem_nodes -= size

                # bfs end
        return ans


        
            
80700