Add solution #1319

Author: JoshCrozier

README.md

@@ -382,6 +382,7 @@
 1313|[Decompress Run-Length Encoded List](./1313-decompress-run-length-encoded-list.js)|Easy|
 1317|[Convert Integer to the Sum of Two No-Zero Integers](./1317-convert-integer-to-the-sum-of-two-no-zero-integers.js)|Easy|
 1318|[Minimum Flips to Make a OR b Equal to c](./1318-minimum-flips-to-make-a-or-b-equal-to-c.js)|Medium|
+1319|[Number of Operations to Make Network Connected](./1319-number-of-operations-to-make-network-connected.js)|Medium|
 1323|[Maximum 69 Number](./1323-maximum-69-number.js)|Easy|
 1324|[Print Words Vertically](./1324-print-words-vertically.js)|Medium|
 1331|[Rank Transform of an Array](./1331-rank-transform-of-an-array.js)|Easy|

solutions/1319-number-of-operations-to-make-network-connected.js

@@ -0,0 +1,50 @@
+/**
+ * 1319. Number of Operations to Make Network Connected
+ * https://leetcode.com/problems/number-of-operations-to-make-network-connected/
+ * Difficulty: Medium
+ *
+ * There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming
+ * a network where connections[i] = [ai, bi] represents a connection between computers ai and bi.
+ * Any computer can reach any other computer directly or indirectly through the network.
+ *
+ * You are given an initial computer network connections. You can extract certain cables between
+ * two directly connected computers, and place them between any pair of disconnected computers to
+ * make them directly connected.
+ *
+ * Return the minimum number of times you need to do this in order to make all the computers
+ * connected. If it is not possible, return -1.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} connections
+ * @return {number}
+ */
+var makeConnected = function(n, connections) {
+  const parent = Array(n).fill(-1);
+  let isNotConnected = n - 1;
+  let count = 0;
+
+  connections.forEach(([node, connection]) => {
+    if (search(node) !== search(connection)) {
+      const p1 = search(node);
+      const p2 = search(connection);
+      if (p1 !== p2) {
+        parent[p2] = p1;
+      }
+      isNotConnected--;
+    } else {
+      count++;
+    }
+  });
+
+  return isNotConnected <= count ? isNotConnected : -1;
+
+  function search(node) {
+    if (parent[node] === -1) {
+      return node;
+    }
+    parent[node] = search(parent[node]);
+    return parent[node];
+  }
+};