Min Cost to Connect All Points
<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1584.Min%20Cost%20to%20Connect%20All%20Points/images/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1584.Min%20Cost%20to%20Connect%20All%20Points/images/c.png" style="width: 214px; height: 268px;" />
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

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

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>
Union Find; Graph; Array; Minimum Spanning Tree
class Solution {
    public int minCostConnectPoints(int[][] points) {
        final int inf = 1 << 30;
        int n = points.length;
        int[][] g = new int[n][n];
        for (int i = 0; i < n; ++i) {
            int x1 = points[i][0], y1 = points[i][1];
            for (int j = i + 1; j < n; ++j) {
                int x2 = points[j][0], y2 = points[j][1];
                int t = Math.abs(x1 - x2) + Math.abs(y1 - y2);
                g[i][j] = t;
                g[j][i] = t;
            }
        }
        int[] dist = new int[n];
        boolean[] vis = new boolean[n];
        Arrays.fill(dist, inf);
        dist[0] = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int j = -1;
            for (int k = 0; k < n; ++k) {
                if (!vis[k] && (j == -1 || dist[k] < dist[j])) {
                    j = k;
                }
            }
            vis[j] = true;
            ans += dist[j];
            for (int k = 0; k < n; ++k) {
                if (!vis[k]) {
                    dist[k] = Math.min(dist[k], g[j][k]);
                }
            }
        }
        return ans;
    }
}