<p>There are <code>n</code> computers numbered from <code>0</code> to <code>n - 1</code> connected by ethernet cables <code>connections</code> forming a network where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a connection between computers <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Any computer can reach any other computer directly or indirectly through the network.</p>

<p>You are given an initial computer network <code>connections</code>. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.</p>

<p>Return <em>the minimum number of times you need to do this in order to make all the computers connected</em>. If it is not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png" style="width: 500px; height: 148px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[0,1],[0,2],[1,2]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Remove cable between computer 1 and 2 and place between computers 1 and 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png" style="width: 500px; height: 129px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are not enough cables.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= connections.length &lt;= min(n * (n - 1) / 2, 10<sup>5</sup>)</code></li>
	<li><code>connections[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 connections.</li>
	<li>No two computers are connected by more than one cable.</li>
</ul>

struct Node {
    int n = 0;
    bool visited = false;
    std::vector<Node*> children;
};

std::pair<int, int> dfs(Node& node) {
    if (node.visited) {
        return {0, 0};
    }

    node.visited = true;
    auto res = std::make_pair(1, node.children.size());
    for (auto child: node.children) {
        auto [nodes, edges] = dfs(*child);
        res.first += nodes;
        res.second += edges;
    }

    return res;
}

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        std::vector<Node> nodes(n);
        for(int i = 0; i < n; ++i) {
            nodes[i].n = i;
            // nodes[i].children.reserve(n);
        }

        for(auto& c: connections) {
            // ptrs still valid because we don't modify nodes vec
            nodes[c[0]].children.push_back(&nodes[c[1]]);
            nodes[c[1]].children.push_back(&nodes[c[0]]);
        }

        int components = 0;
        int remains = 0;
        for (auto& n: nodes) {
            if (!n.visited) {
                components += 1;
                auto [nodes_num, edges_num] = dfs(n);
                if (edges_num % 2 == 1) {
                    throw std::runtime_error("Unexpected number of edges");
                }
                remains += (edges_num / 2) - (nodes_num - 1);
            }
        }
        std::cout << components << ' ' << remains << std::endl;
        if (components - 1 > remains) {
            return -1;
        } else {
            return components - 1;
        }
    }
};
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