Second Minimum Time to Reach Destination

class Solution {
public:
    int findSecondMinTime(int nodes, vector<vector<int>>& links, int travelTime, int signalChange) {
        vector<vector<int>> graph(nodes + 1);
        for (auto& link : links) {
            graph[link[0]].push_back(link[1]);
            graph[link[1]].push_back(link[0]);
        }
    
        queue<pair<int, int>> explorationQueue;
        vector<int> firstVisit(nodes + 1, -1), secondVisit(nodes + 1, -1);
        explorationQueue.push({1, 1});
        firstVisit[1] = 0;
    
        while (!explorationQueue.empty()) {
            auto [currentNode, visitCount] = explorationQueue.front();
            explorationQueue.pop();
    
            int currentDuration = visitCount == 1 ? firstVisit[currentNode] : secondVisit[currentNode];
            if ((currentDuration / signalChange) % 2) {
                currentDuration = signalChange * (currentDuration / signalChange + 1) + travelTime;
            } else {
                currentDuration += travelTime;
            }
    
            for (int adjacent : graph[currentNode]) {
                if (firstVisit[adjacent] == -1) {
                    firstVisit[adjacent] = currentDuration;
                    explorationQueue.push({adjacent, 1});
                } else if (secondVisit[adjacent] == -1 && firstVisit[adjacent] != currentDuration) {
                    if (adjacent == nodes) return currentDuration;
                    secondVisit[adjacent] = currentDuration;
                    explorationQueue.push({adjacent, 2});
                }
            }
        }
        return -1;
    }
};

This C++ solution determines the second minimum time required to reach the last node from the first node in a network of nodes interconnected by edges.

• Define a class Solution with a public function findSecondMinTime that accepts four parameters:

  • nodes: total number of nodes.
  • links: vector of pairs representing bidirectional connections between nodes.
  • travelTime: time taken to travel between each connected node pair.
  • signalChange: time after which traffic signals change.

• Represent the network as an adjacency list using a vector of vectors, graph, where each index represents a node and stores its interconnected nodes.

• Use a queue of pairs, explorationQueue, for the BFS mechanism. Each pair in the queue represents a node and a visit count (first or second visit to the node).

• Use two vectors, firstVisit and secondVisit, initialized with -1, to track the time of the first and second visits to each node, respectively.

• Initialize the queue by pushing the starting node with a visitCount of 1 and set its firstVisit to 0 (start time).

• Within the BFS loop:

  • Dequeue the front pair (node and visit count) and calculate current travel currentDuration based on the node’s visit time.
  • Adjust currentDuration by checking traffic signals' change time. If in the middle of traffic change, wait until signals allow passage, then add the travelTime.
  • Traverse through the adjacent nodes:
    • For a node visited the first time, set firstVisit to currentDuration and queue it with a visitCount of 1.
    • For a node visited the second time and if the current duration differs from the first visit, check if it's the destination node. If so, return the currentDuration. Otherwise, set secondVisit and enqueue it with a visitCount of 2.

• If no path provides a second visit scenario by the end of the BFS, return -1.

This approach effectively calculates and tracks two distinct travel times for each node considering the constraints of signal changes, thereby finding the second shortest time to the destination.

class Solution {
    
    public int findSecondMinTime(int nodes, int[][] links, int travelTime, int signalChange) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] link : links) {
            int src = link[0], dest = link[1];
            graph.computeIfAbsent(src, k -> new ArrayList<>()).add(dest);
            graph.computeIfAbsent(dest, k -> new ArrayList<>()).add(src);
        }
        int[] firstVisit = new int[nodes + 1], secondVisit = new int[nodes + 1];
        Arrays.fill(firstVisit, -1);
        Arrays.fill(secondVisit, -1);
        Queue<int[]> bfsQueue = new LinkedList<>();
        bfsQueue.add(new int[] { 1, 1 });
        firstVisit[1] = 0;
    
        while (!bfsQueue.isEmpty()) {
            int[] current = bfsQueue.poll();
            int currentNode = current[0];
            int occurrence = current[1];
    
            int currentTravelTime = occurrence == 1 ? firstVisit[currentNode] : secondVisit[currentNode];
            // Determine wait times based on signal timing
            if ((currentTravelTime / signalChange) % 2 != 0) {
                currentTravelTime = signalChange * (currentTravelTime / signalChange + 1) + travelTime;
            } else {
                currentTravelTime += travelTime;
            }
    
            for (int adjacentNode : graph.get(currentNode)) {
                if (firstVisit[adjacentNode] == -1) {
                    firstVisit[adjacentNode] = currentTravelTime;
                    bfsQueue.add(new int[] { adjacentNode, 1 });
                } else if (
                    secondVisit[adjacentNode] == -1 && firstVisit[adjacentNode] != currentTravelTime
                ) {
                    if (adjacentNode == nodes) return currentTravelTime;
                    secondVisit[adjacentNode] = currentTravelTime;
                    bfsQueue.add(new int[] { adjacentNode, 2 });
                }
            }
        }
        return -1;
    }
}

This Java program provides a solution to find the second shortest time to reach the final destination in a graph. The graph nodes represent locations with edges defined as travel links between them. Given specific travel and signal conditions, this solution effectively utilizes BFS (Breadth-First Search) to explore all possible routes.

• Define a graph using a HashMap to store connections between nodes.
• Utilize two arrays, firstVisit and secondVisit, to track the first and second times a node is visited.
• Employ a queue mechanism (bfsQueue) to explore nodes level-by-level (BFS approach).
• Ensure time adjustments based on the signal change, making a node accessible only when the signal allows.
• Process each node by exploring its neighbors and calculating their respective travel times, updating the firstVisit and secondVisit arrays accordingly.
• If the node has been visited once, attempt to visit it a second time with a different timing to potentially calculate a different route.
• Return the travel time as soon as the second visit time is found for the final node; otherwise, return -1 if no such path exists.

Key considerations include:

• Managing signal timings to determine when a node can be accessed, critical in evaluating the correct next travel time.
• Exploring alternate paths efficiently by immediately stopping the search when the second shortest path to the destination is identified.
• Utilizing BFS ensures that the paths evaluated are among the shortest possible before considering alternate, longer paths due to the layered nature of BFS.

class Solution:
    def findSecondMinimumTime(self, totalNodes, connections, travelTime, signalChange):
        from collections import defaultdict, deque
    
        graph = defaultdict(list)
        for src, dst in connections:
            graph[src].append(dst)
            graph[dst].append(src)
    
        firstVisitTime, secondVisitTime = [-1] * (totalNodes + 1), [-1] * (totalNodes + 1)
        q = deque([(1, 1)])
        firstVisitTime[1] = 0
    
        while q:
            current, frequency = q.popleft()
            current_time = firstVisitTime[current] if frequency == 1 else secondVisitTime[current]
    
            if (current_time // signalChange) % 2 == 1:
                current_time = (current_time // signalChange + 1) * signalChange + travelTime
            else:
                current_time += travelTime
    
            for adjacent in graph[current]:
                if firstVisitTime[adjacent] == -1:
                    firstVisitTime[adjacent] = current_time
                    q.append((adjacent, 1))
                elif secondVisitTime[adjacent] == -1 and firstVisitTime[adjacent] != current_time:
                    if adjacent == totalNodes:
                        return current_time
                    secondVisitTime[adjacent] = current_time
                    q.append((adjacent, 2))
    
        return 0

This summary explains an algorithm implemented in Python for finding the second minimum time to reach the final destination in a traffic network. The network is represented as an undirected graph where nodes correspond to intersections and edges correspond to roads between these intersections.

• Begin by creating a graph representation to model the network. This involves constructing an adjacency list for each node to identify directly connected nodes.
• Utilize a deque from the collections module for efficient population and retrieval of nodes to visit.
• Set up two lists firstVisitTime and secondVisitTime to store times when each node is visited for the first and second time respectively. Use -1 to represent unvisited nodes.
• Start the traversal from node 1, marking it with an initial travel time of 0.
• While traversing:
  • Calculate the current time of arrival at each node considering whether a signal change is needed and add the standard travel time.
  • Iterate through adjacent nodes:
    • Update the first visited time if the node hasn't been visited.
    • For the second visited time, check that the time is different from when it was first visited. If conditions are met and the node is the final destination, return the current time.
  • Handle signal frequency to ensure traversal at allowed intervals based on the signalChange time.
• If traversal is completed and the destination was visited twice, or conditions to capture second visit are not met, return 0.

This implementation leverages BFS (Breadth-first Search) enhanced by checking travel times and handling signal changes to ensure the accurate calculation of the shortest and second-shortest times to reach the destination under given constraints.