<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>

#define ll long long
#include <vector>
#include <unordered_map>

using namespace std;

class DisjointSetUnion {
private:
    vector<ll> parent, componentSize;
public:
    DisjointSetUnion(ll n) {  // Changed parameter name to n
        parent.resize(n);
        componentSize.resize(n, 1); // Initialize sizes to 1 for each component
        for (ll i = 0; i < n; i++)
            parent[i] = i;
    }

    ll findParent(ll u) {
        if (parent[u] != u)
            parent[u] = findParent(parent[u]); // Path compression
        return parent[u];
    }

    void unionBySize(ll u, ll v) {
        ll rootU = findParent(u);
        ll rootV = findParent(v);

        if (rootU != rootV) {
            // Attach smaller tree under the larger tree
            if (componentSize[rootU] < componentSize[rootV]) {
                parent[rootU] = rootV;
                componentSize[rootV] += componentSize[rootU];
            } else {
                parent[rootV] = rootU;
                componentSize[rootU] += componentSize[rootV];
            }
        }
    }

    vector<ll> getComponentSizes() {
        vector<ll> sizes(parent.size(), 0);
        for (ll i = 0; i < parent.size(); i++) {
            ll root = findParent(i);
            sizes[root] = componentSize[root];
        }
        return sizes;
    }
};

class Solution {
public:
    long long countPairs(ll n, vector<vector<int>>& edges) {
        DisjointSetUnion dsu(n);

        for (auto& edge : edges) {
            ll u = edge[0], v = edge[1];
            dsu.unionBySize(u, v);
        }

        // Calculate size of each component
        vector<ll> componentSizes = dsu.getComponentSizes();

        // Count the number of pairs
        ll totalPairs = 0;
        // ll sum = 0;
        // for (ll size : componentSizes) {
        //     if (size > 0) {
        //         totalPairs += size * sum;
        //         sum += size;
        //     }
        // }

        ll totalComponents = n;
        for(auto &componentSize : componentSizes){
            if(componentSize > 0){
                totalPairs += componentSize * (totalComponents - componentSize);
                totalComponents -= componentSize;
            }
        }

        return totalPairs;
    }
};

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