711400. Number of Ways to Arrive at Destination
Number of Ways to Arrive at Destination
Summary
The problem is about finding the number of ways a person can arrive at their destination given a set of paths with varying costs. The key concept involved is shortest path, which is typically solved using algorithms such as Dijkstra's or Bellman-Ford.
Detailed Explanation
To solve this problem, we will use a graph-based approach. We'll create a weighted graph where each node represents a location and the edges represent the paths between locations. Each edge will have a weight (cost) associated with it.
We can use Dijkstra's algorithm to find the shortest path from the starting point to all other points in the graph, which represents the minimum cost required to arrive at each destination. The number of ways to arrive at a destination is then equal to the total number of paths that have the same minimum cost as the shortest path.
Here's a step-by-step breakdown of the solution:
- Create a weighted graph with nodes representing locations and edges representing paths between locations.
- Use Dijkstra's algorithm to find the shortest path from the starting point to all other points in the graph.
- For each destination, count the number of paths that have the same minimum cost as the shortest path.
Time complexity: O(E + V log V), where E is the number of edges and V is the number of vertices. This is because Dijkstra's algorithm has a time complexity of O(E + V log V).
Space complexity: O(V), where V is the number of vertices. This is because we need to store the shortest distances and previous nodes for each vertex.
Optimized Solutions
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
// Create a graph with weighted edges
Map<String, Map<String, Integer>> graph = new HashMap<>();
graph.put("A", new HashMap<>());
graph.get("A").put("B", 2);
graph.get("A").put("C", 3);
graph.put("B", new HashMap<>());
graph.get("B").put("D", 1);
graph.put("C", new HashMap<>());
graph.get("C").put("D", 2);
// Find the shortest paths using Dijkstra's algorithm
Map<String, Integer> distances = new HashMap<>();
Map<String, String> previousNodes = new HashMap<>();
dijkstra(graph, "A", distances, previousNodes);
// Count the number of ways to arrive at each destination
int totalWays = 0;
for (Map.Entry<String, Integer> entry : distances.entrySet()) {
if (entry.getValue() == getMinimumCost(graph, "A")) {
totalWays++;
}
}
System.out.println("Total number of ways: " + totalWays);
}
public static void dijkstra(Map<String, Map<String, Integer>> graph, String startNode, Map<String, Integer> distances, Map<String, String> previousNodes) {
Set<String> visited = new HashSet<>();
PriorityQueue<Map.Entry<String, Integer>> queue = new PriorityQueue<>();
// Initialize the distances and previous nodes
for (Map.Entry<String, Map<String, Integer>> entry : graph.entrySet()) {
distances.put(entry.getKey(), Integer.MAX_VALUE);
previousNodes.put(entry.getKey(), null);
}
distances.put(startNode, 0);
// Run Dijkstra's algorithm
queue.add(new AbstractMap.SimpleEntry<>(startNode, 0));
while (!queue.isEmpty()) {
Map.Entry<String, Integer> current = queue.poll();
if (visited.contains(current.getKey())) continue;
visited.add(current.getKey());
for (Map.Entry<String, Integer> edge : graph.get(current.getKey()).entrySet()) {
int newDistance = distances.get(current.getKey()) + edge.getValue();
if (newDistance < distances.getOrDefault(edge.getKey(), Integer.MAX_VALUE)) {
distances.put(edge.getKey(), newDistance);
previousNodes.put(edge.getKey(), current.getKey());
queue.add(new AbstractMap.SimpleEntry<>(edge.getKey(), newDistance));
}
}
}
}
public static int getMinimumCost(Map<String, Map<String, Integer>> graph, String startNode) {
return distances.getOrDefault(startNode, Integer.MAX_VALUE);
}
}