Add solution #1615

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,421 LeetCode solutions in JavaScript
+# 1,422 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1245,6 +1245,7 @@
 1610|[Maximum Number of Visible Points](./solutions/1610-maximum-number-of-visible-points.js)|Hard|
 1611|[Minimum One Bit Operations to Make Integers Zero](./solutions/1611-minimum-one-bit-operations-to-make-integers-zero.js)|Hard|
 1614|[Maximum Nesting Depth of the Parentheses](./solutions/1614-maximum-nesting-depth-of-the-parentheses.js)|Easy|
+1615|[Maximal Network Rank](./solutions/1615-maximal-network-rank.js)|Medium|
 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|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1615-maximal-network-rank.js

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+/**
+ * 1615. Maximal Network Rank
+ * https://leetcode.com/problems/maximal-network-rank/
+ * Difficulty: Medium
+ *
+ * There is an infrastructure of n cities with some number of roads connecting these cities.
+ * Each roads[i] = [ai, bi] indicates that there is a bidirectional road between cities ai and bi.
+ *
+ * The network rank of two different cities is defined as the total number of directly connected
+ * roads to either city. If a road is directly connected to both cities, it is only counted once.
+ *
+ * The maximal network rank of the infrastructure is the maximum network rank of all pairs of
+ * different cities.
+ *
+ * Given the integer n and the array roads, return the maximal network rank of the entire
+ * infrastructure.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} roads
+ * @return {number}
+ */
+var maximalNetworkRank = function(n, citiesConnections) {
+  const cityRoadCounts = new Array(n).fill(0);
+  const directConnections = new Set();
+
+  for (const [cityA, cityB] of citiesConnections) {
+    cityRoadCounts[cityA]++;
+    cityRoadCounts[cityB]++;
+    directConnections.add(`${Math.min(cityA, cityB)}-${Math.max(cityA, cityB)}`);
+  }
+
+  let maxNetworkRank = 0;
+
+  for (let cityA = 0; cityA < n; cityA++) {
+    for (let cityB = cityA + 1; cityB < n; cityB++) {
+      let networkRank = cityRoadCounts[cityA] + cityRoadCounts[cityB];
+      if (directConnections.has(`${cityA}-${cityB}`)
+          || directConnections.has(`${cityB}-${cityA}`)) {
+        networkRank--;
+      }
+      maxNetworkRank = Math.max(maxNetworkRank, networkRank);
+    }
+  }
+
+  return maxNetworkRank;
+};