<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://assets.leetcode.com/uploads/2020/08/26/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://assets.leetcode.com/uploads/2020/08/26/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>

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        // Create adjacency list
        unordered_map<int, list<pair<int, int>>> adj;
        int n = points.size();

        // Build the adjacency list by calculating Manhattan distances
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int u = i;
                int v = j;
                int w = abs(points[u][0] - points[v][0]) + abs(points[u][1] - points[v][1]);

                adj[u].push_back({v, w});
                adj[v].push_back({u, w}); // Undirected graph
            }
        }

        // Using unordered_map instead of vector
        unordered_map<int, bool> vis;   // To mark visited nodes
        unordered_map<int, int> key;    // To store minimum weights for each node
        unordered_map<int, int> parent; // To store the parent node of each node

        // Initialize all keys as INT_MAX and parent as -1
        for (int i = 0; i < n; i++) {
            key[i] = INT_MAX;
            parent[i] = -1;
            vis[i] = false;  // Initially, no node is visited
        }

        // Prim's algorithm starts, key of the starting node (0) is set to 0
        key[0] = 0;

        // Iterate n times to construct the minimum spanning tree (MST)
        for (int i = 0; i < n; i++) {
            int mini = INT_MAX;
            int u = -1;

            // Find the node with the minimum key value that hasn't been visited
            for (int v = 0; v < n; v++) {
                if (!vis[v] && key[v] < mini) {
                    mini = key[v];
                    u = v;
                }
            }

            if (u == -1) {
                return -1; // Graph is not connected
            }

            // Mark the node as visited
            vis[u] = true;

            // Update the key and parent for the neighboring nodes
            for (auto neighbour : adj[u]) {
                int v = neighbour.first;
                int w = neighbour.second;

                if (!vis[v] && w < key[v]) {
                    key[v] = w;
                    parent[v] = u;
                }
            }
        }

        int ans = 0;
        // Sum up the key values to get the total minimum cost
        for (int i = 0; i < n; i++) {
            ans += key[i];
        }
        return ans;
    }
};

215565