Daily Solutions With JS

Author: bola-nabil

Hard/2045. Second Minimum Time to Reach Destination.js

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+const initializeGraph = (n) => { let G = []; for (let i = 0; i < n; i++) { G.push([]); } return G; };
+const addEdgeToG = (G, Edges) => { for (const [u, v] of Edges) { G[u].push(v); G[v].push(u); } };
+
+const secondMinimum = (n, edges, time, change) => {
+    let adj = initializeGraph(n + 1);
+    addEdgeToG(adj, edges);
+    let cost = initializeGraph(n + 1);
+    let pq = new MinPriorityQueue({ priority: x => x[0] });
+    pq.enqueue([0, 1]);
+    let green = 2 * change;
+    while (pq.size()) {
+        let cur = pq.dequeue().element;
+        let [t, node] = cur;
+        if (cost[node].length == 2) continue;
+        let nextT = t % green < change ? t : ((t + green - 1) / green >> 0) * green;
+        let cn = cost[node].length;
+        if (node == n) {
+            if (cn == 0 || cost[node][cn - 1] != t) {
+                cost[node].push(t);
+            } else {
+                continue;
+            }
+        } else {
+            if (cn == 0 || cost[node][cn - 1] != nextT) {
+                cost[node].push(nextT);
+            } else {
+                continue;
+            }
+        }
+        for (const next_node of adj[node]) pq.enqueue([nextT + time, next_node]);
+    }
+    return cost[n][1];
+};