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Problem solution · Java

Minimum Cost to Reach Destination in Time

Minimum Cost to Reach Destination in Time: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
59 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Minimum Cost to Reach Destination in Time, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 59 lines of Java from the credited upstream file 1928.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 3 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Cost to Reach Destination in Time · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minCost(int maxTime, int[][] edges, int[] passingFees) {    final int n = passingFees.length;    List<Pair<Integer, Integer>>[] graph = new List[n];    Arrays.setAll(graph, i -> new ArrayList<>());     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      final int t = edge[2];      graph[u].add(new Pair<>(v, t));      graph[v].add(new Pair<>(u, t));    }     return dijkstra(graph, 0, n - 1, maxTime, passingFees);  }   private int dijkstra(List<Pair<Integer, Integer>>[] graph, int src, int dst, int maxTime,                       int[] passingFees) {    int[] cost = new int[graph.length]; // cost[i] := the minimum cost to reach the i-th city    int[] dist = new int[graph.length]; // dist[i] := the minimum distance to reach the i-th city    Arrays.fill(cost, Integer.MAX_VALUE);    Arrays.fill(dist, maxTime + 1);     cost[0] = passingFees[0];    dist[0] = 0;    Queue<int[]> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])) {      { offer(new int[] {cost[src], dist[src], src}); } // (cost[u], dist[u], u)    };     while (!minHeap.isEmpty()) {      final int currCost = minHeap.peek()[0];      final int d = minHeap.peek()[1];      final int u = minHeap.poll()[2];      if (u == dst)        return cost[dst];      if (d > dist[u] && currCost > cost[u])        continue;      for (Pair<Integer, Integer> pair : graph[u]) {        final int v = pair.getKey();        final int w = pair.getValue();        if (d + w > maxTime)          continue;        // Go from x -> y        if (currCost + passingFees[v] < cost[v]) {          cost[v] = currCost + passingFees[v];          dist[v] = d + w;          minHeap.offer(new int[] {cost[v], dist[v], v});        } else if (d + w < dist[v]) {          dist[v] = d + w;          minHeap.offer(new int[] {currCost + passingFees[v], dist[v], v});        }      }    }     return -1;  }}

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