Add solution #2045

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,707 LeetCode solutions in JavaScript
+# 1,708 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1568,6 +1568,7 @@
 2042|[Check if Numbers Are Ascending in a Sentence](./solutions/2042-check-if-numbers-are-ascending-in-a-sentence.js)|Easy|
 2043|[Simple Bank System](./solutions/2043-simple-bank-system.js)|Medium|
 2044|[Count Number of Maximum Bitwise-OR Subsets](./solutions/2044-count-number-of-maximum-bitwise-or-subsets.js)|Medium|
+2045|[Second Minimum Time to Reach Destination](./solutions/2045-second-minimum-time-to-reach-destination.js)|Hard|
 2047|[Number of Valid Words in a Sentence](./solutions/2047-number-of-valid-words-in-a-sentence.js)|Easy|
 2053|[Kth Distinct String in an Array](./solutions/2053-kth-distinct-string-in-an-array.js)|Medium|
 2071|[Maximum Number of Tasks You Can Assign](./solutions/2071-maximum-number-of-tasks-you-can-assign.js)|Hard|

solutions/2045-second-minimum-time-to-reach-destination.js

@@ -0,0 +1,67 @@
+/**
+ * 2045. Second Minimum Time to Reach Destination
+ * https://leetcode.com/problems/second-minimum-time-to-reach-destination/
+ * Difficulty: Hard
+ *
+ * A city is represented as a bi-directional connected graph with n vertices where each vertex
+ * is labeled from 1 to n (inclusive). The edges in the graph are represented as a 2D integer
+ * array edges, where each edges[i] = [ui, vi] denotes a bi-directional edge between vertex
+ * ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has
+ * an edge to itself. The time taken to traverse any edge is time minutes.
+ *
+ * Each vertex has a traffic signal which changes its color from green to red and vice versa
+ * every change minutes. All signals change at the same time. You can enter a vertex at any
+ * time, but can leave a vertex only when the signal is green. You cannot wait at a vertex
+ * if the signal is green.
+ *
+ * The second minimum value is defined as the smallest value strictly larger than the minimum value.
+ * - For example the second minimum value of [2, 3, 4] is 3, and the second minimum value of
+ *   [2, 2, 4] is 4.
+ *
+ * Given n, edges, time, and change, return the second minimum time it will take to go from
+ * vertex 1 to vertex n.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @param {number} time
+ * @param {number} change
+ * @return {number}
+ */
+var secondMinimum = function(n, edges, time, change) {
+  const adjList = Array.from({ length: n + 1 }, () => []);
+  for (const [u, v] of edges) {
+    adjList[u].push(v);
+    adjList[v].push(u);
+  }
+
+  const distances = Array.from({ length: n + 1 }, () => [Infinity, Infinity]);
+  const queue = [[1, 0]];
+  distances[1][0] = 0;
+
+  while (queue.length) {
+    const [node, currTime] = queue.shift();
+
+    for (const next of adjList[node]) {
+      const signalCycle = Math.floor(currTime / change);
+      const isGreen = signalCycle % 2 === 0;
+      let nextTime = currTime + time;
+
+      if (!isGreen) {
+        nextTime = (signalCycle + 1) * change + time;
+      }
+
+      if (nextTime < distances[next][0]) {
+        distances[next][1] = distances[next][0];
+        distances[next][0] = nextTime;
+        queue.push([next, nextTime]);
+      } else if (nextTime > distances[next][0] && nextTime < distances[next][1]) {
+        distances[next][1] = nextTime;
+        queue.push([next, nextTime]);
+      }
+    }
+  }
+
+  return distances[n][1];
+};