<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>

class dsj {
    public:
    vector<int> parent , rank;
    int n;

    dsj(int v) {
        n = v;
        for(int i = 0;i<v;i++) {
            parent.push_back(i);
            rank.push_back(0);
        }
    }

    int findParent(int x) {
        if(x == parent[x]) {
            return x;
        }

        return parent[x] = findParent(parent[x]);
    }

    void unionf(int u , int v) {
        int ulpU = findParent(u) , ulpV = findParent(v);

        if(ulpU == ulpV) {
            return;
        }

        if(rank[ulpU] < rank[ulpV]) {
            parent[ulpU] = ulpV;
        } else if(rank[ulpU] > rank[ulpV]) {
            parent[ulpV] = ulpU;
        } else {
            parent[ulpV] = ulpU;
            rank[ulpU]++;
        }
    }
};

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        dsj ds(n);

        for(auto i : connections) {
            ds.unionf(i[0] , i[1]);
        }

        unordered_set<int> st;

        for(auto i : ds.parent) { // count distinct parents
            st.insert(ds.findParent(i));
        }

        // st.size() is number of components
        if(connections.size() < n - 1) {
            return -1;
        }

        return st.size() - 1;
    }
};
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