Number of Ways to Arrive at Destination problem solved

Author: hasibulislam999

Shortest Path/1976_number-of-ways-to-arrive-at-destination.js

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+/**
+ * Title: Number of Ways to Arrive at Destination
+ * Description: You are in a city that consists of n intersections numbered from 0 to n - 1 with bi-directional roads between some intersections. The inputs are generated such that you can reach any intersection from any other intersection and that there is at most one road between any two intersections.
+ * Author: Hasibul Islam
+ * Date: 04/05/2023
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} roads
+ * @return {number}
+ */
+const mod = 1e9 + 7;
+const countPaths = (n, road) => {
+  let adj = initializeGraph(n);
+  for (const [u, v, cost] of road) {
+    adj[u].push([v, cost]);
+    adj[v].push([u, cost]);
+  }
+  return dijkstra(n, adj, 0);
+};
+
+const dijkstra = (n, g, source) => {
+  // g: adjacent graph list, n: total vertices
+  let dist = Array(n).fill(Number.MAX_SAFE_INTEGER);
+  let ways = Array(n).fill(0);
+  const pq = new MinPriorityQueue({ priority: (x) => x[0] * 200 + x[1] });
+  dist[0] = 0;
+  ways[0] = 1;
+  pq.enqueue([0, source]);
+  while (pq.size()) {
+    let cur = pq.dequeue().element;
+    let [curCost, curNode] = cur;
+    if (dist[curNode] != curCost) continue;
+    for (const [node, cost] of g[curNode]) {
+      // parse neighbour node
+      let newDis = curCost + cost;
+      if (newDis == dist[node]) {
+        ways[node] += ways[curNode];
+        ways[node] %= mod;
+      } else if (newDis < dist[node]) {
+        dist[node] = newDis;
+        ways[node] = ways[curNode];
+        pq.enqueue([dist[node], node]);
+      }
+    }
+  }
+  return ways[n - 1];
+};
+
+const initializeGraph = (n) => {
+  let G = [];
+  for (let i = 0; i < n; i++) {
+    G.push([]);
+  }
+  return G;
+};