Add solution #1584

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,384 LeetCode solutions in JavaScript
+# 1,385 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1209,6 +1209,7 @@
 1579|[Remove Max Number of Edges to Keep Graph Fully Traversable](./solutions/1579-remove-max-number-of-edges-to-keep-graph-fully-traversable.js)|Hard|
 1582|[Special Positions in a Binary Matrix](./solutions/1582-special-positions-in-a-binary-matrix.js)|Easy|
 1583|[Count Unhappy Friends](./solutions/1583-count-unhappy-friends.js)|Medium|
+1584|[Min Cost to Connect All Points](./solutions/1584-min-cost-to-connect-all-points.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|

solutions/1584-min-cost-to-connect-all-points.js

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+/**
+ * 1584. Min Cost to Connect All Points
+ * https://leetcode.com/problems/min-cost-to-connect-all-points/
+ * Difficulty: Medium
+ *
+ * You are given an array points representing integer coordinates of some points on a 2D-plane,
+ * where points[i] = [xi, yi].
+ *
+ * The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between
+ * them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.
+ *
+ * Return the minimum cost to make all points connected. All points are connected if there is
+ * exactly one simple path between any two points.
+ */
+
+/**
+ * @param {number[][]} points
+ * @return {number}
+ */
+var minCostConnectPoints = function(points) {
+  const n = points.length;
+  const minCost = Array(n).fill(Infinity);
+  const visited = new Set();
+  let result = 0;
+
+  minCost[0] = 0;
+
+  for (let i = 0; i < n; i++) {
+    let minIdx = -1;
+    let minVal = Infinity;
+
+    for (let j = 0; j < n; j++) {
+      if (!visited.has(j) && minCost[j] < minVal) {
+        minVal = minCost[j];
+        minIdx = j;
+      }
+    }
+
+    if (minIdx === -1) break;
+
+    visited.add(minIdx);
+    result += minVal;
+
+    for (let j = 0; j < n; j++) {
+      if (!visited.has(j)) {
+        const cost = Math.abs(points[minIdx][0] - points[j][0])
+          + Math.abs(points[minIdx][1] - points[j][1]);
+        minCost[j] = Math.min(minCost[j], cost);
+      }
+    }
+  }
+
+  return result;
+};