Connecting Cities With Minimum Cost
<p>There are <code>n</code> cities labeled from <code>1</code> to <code>n</code>. You are given the integer <code>n</code> and an array <code>connections</code> where <code>connections[i] = [x<sub>i</sub>, y<sub>i</sub>, cost<sub>i</sub>]</code> indicates that the cost of connecting city <code>x<sub>i</sub></code> and city <code>y<sub>i</sub></code> (bidirectional connection) is <code>cost<sub>i</sub></code>.</p>

<p>Return <em>the minimum <strong>cost</strong> to connect all the </em><code>n</code><em> cities such that there is at least one path between each pair of cities</em>. If it is impossible to connect all the <code>n</code> cities, return <code>-1</code>,</p>

<p>The <strong>cost</strong> is the sum of the connections&#39; costs used.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1135.Connecting%20Cities%20With%20Minimum%20Cost/images/1314_ex2.png" style="width: 161px; height: 141px;" />
<pre>
<strong>Input:</strong> n = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Choosing any 2 edges will connect all cities so we choose the minimum 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1135.Connecting%20Cities%20With%20Minimum%20Cost/images/1314_ex1.png" style="width: 136px; height: 91px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[1,2,3],[3,4,4]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no way to connect all cities even if all edges are used.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= connections.length &lt;= 10<sup>4</sup></code></li>
	<li><code>connections[i].length == 3</code></li>
	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= n</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li><code>0 &lt;= cost<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
</ul>
Union Find; Graph; Minimum Spanning Tree; Heap (Priority Queue)
class Solution {
public:
    int minimumCost(int n, vector<vector<int>>& connections) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        sort(connections.begin(), connections.end(), [](auto& a, auto& b) { return a[2] < b[2]; });
        int ans = 0;
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        for (auto& e : connections) {
            int x = e[0] - 1, y = e[1] - 1, cost = e[2];
            if (find(x) == find(y)) {
                continue;
            }
            p[find(x)] = find(y);
            ans += cost;
            if (--n == 1) {
                return ans;
            }
        }
        return -1;
    }
};