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Second Minimum Time to Reach Destination

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Graphs

A city is represented as a bi-directional connected graph with n vertices where each vertex is labeled from 1 to n (inclusive). The edges in the graph are represented as a 2D integer array edges, where each edges[i] = [u_i, v_i] denotes a bi-directional edge between vertex u_i and vertex v_i. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself. The time taken to traverse any edge is time minutes.

Each vertex has a traffic signal which changes its color from green to red and vice versa every change minutes. All signals change at the same time. You can enter a vertex at any time, but can leave a vertex only when the signal is green. You cannot wait at a vertex if the signal is green.

The second minimum value is defined as the smallest value strictly larger than the minimum value.

  • For example the second minimum value of [2, 3, 4] is 3, and the second minimum value of [2, 2, 4] is 4.

Given n, edges, time, and change, return the second minimum time it will take to go from vertex 1 to vertex n.

Notes:

  • You can go through any vertex any number of times, including 1 and n.
  • You can assume that when the journey starts, all signals have just turned green.

Example 1:

n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5

Output: 13

Explanation: The figure on the left shows the given graph. The blue path in the figure on the right is the minimum time path. The time taken is:

  • Start at 1, time elapsed=0
  • 1 -> 4: 3 minutes, time elapsed=3
  • 4 -> 5: 3 minutes, time elapsed=6

Hence the minimum time needed is 6 minutes.

The red path shows the path to get the second minimum time.

  • Start at 1, time elapsed=0
  • 1 -> 3: 3 minutes, time elapsed=3
  • 3 -> 4: 3 minutes, time elapsed=6
  • Wait at 4 for 4 minutes, time elapsed=10
  • 4 -> 5: 3 minutes, time elapsed=13

Hence the second minimum time is 13 minutes.

Example 2:

n = 2, edges = [[1,2]], time = 3, change = 2

Output: 11

Explanation: The minimum time path is 1 -> 2 with time = 3 minutes. The second minimum time path is 1 -> 2 -> 1 -> 2 with time = 11 minutes.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the number of nodes (n) and the number of edges in the graph? Can n be as large as 10^5 or 10^6?
  2. What are the possible values for `time` and `change`? Are they always positive integers, and are there any specific upper bounds I should be aware of?
  3. If there is no path that reaches the destination with exactly the second minimum time, what should I return? For example, should I return -1 or a specific error value?
  4. Can there be multiple edges between two nodes, and can the graph be disconnected?
  5. What does 'second minimum time' mean precisely? If there are multiple paths with the same minimum time, is the second minimum still the next smallest time *after* the minimum time, or the time of a *different* path that takes longer than the absolute minimum?

Brute Force Solution

Approach

We want to find the second fastest way to get from one place to another in a city. The brute force method essentially tries every possible route, keeping track of how long each one takes.

Here's how the algorithm would work step-by-step:

  1. Start by listing every possible path from the starting place to the destination.
  2. Calculate the travel time for each of these paths, taking into account any waiting time at traffic lights.
  3. After computing the time for every possible path, sort all the times from shortest to longest.
  4. The shortest time represents the fastest route. The next shortest time, if it's different from the shortest time, represents the second fastest route.

Code Implementation

def second_minimum_time_brute_force(number_of_nodes, edges, time_to_travel, change):
    # Generate all possible paths from node 1 to node n.
    all_paths = find_all_paths(number_of_nodes, edges, 1, number_of_nodes)

    path_times = []
    for path in all_paths:
        current_time = 0
        for i in range(len(path) - 1):
            current_time += time_to_travel
            # Simulate traffic lights: wait if needed.
            light_phase = (current_time // change) % 2
            if light_phase == 1:
                current_time += change - (current_time % change)
        path_times.append(current_time)

    # Sort all travel times to find the minimum and second minimum.
    path_times.sort()

    minimum_time = path_times[0]

    # Find the second minimum time, ensuring it's different from the minimum.
    for time in path_times:
        if time > minimum_time:
            return time

    return -1

def find_all_paths(number_of_nodes, edges, start_node, destination_node):
    graph = {i: [] for i in range(1, number_of_nodes + 1)}
    for source, destination in edges:
        graph[source].append(destination)
        graph[destination].append(source)

    paths = []
    
    def dfs(current_node, current_path):
        current_path.append(current_node)
        if current_node == destination_node:
            paths.append(current_path.copy())
        else:
            for neighbor in graph[current_node]:
                if len(current_path) < 2 * number_of_nodes: #avoid cycles
                    dfs(neighbor, current_path)
        current_path.pop()

    dfs(start_node, [])
    return paths

Big(O) Analysis

Time Complexity
O(V!)The brute force approach involves listing every possible path from the start to the destination in a graph of V vertices and E edges. In the worst-case scenario, where the graph is dense and highly interconnected, the number of possible paths can grow factorially with the number of vertices (V!). Calculating the travel time for each path takes O(V) time in the worst case, as a path could visit all vertices. Sorting the path times also incurs additional time complexity, but it is dominated by the path enumeration and is negligible compared to the factorial growth of paths generated. The total time complexity is therefore bounded by O(V * V!) (path calculation) which is approximated by O(V!).
Space Complexity
O(N!)The brute force method lists every possible path from the starting place to the destination. In the worst-case scenario, this could involve enumerating all permutations of nodes in the graph, leading to a list of paths with a size proportional to N! where N represents the number of nodes (places) in the city. Storing these paths and their corresponding travel times requires auxiliary space proportional to the number of paths considered, hence O(N!). Sorting these times in step 3 does not increase the already massive space complexity.

Optimal Solution

Approach

The goal is to find the second shortest time to reach the destination. We use a modified version of the shortest path algorithm, keeping track of both the shortest and second shortest times to reach each location.

Here's how the algorithm would work step-by-step:

  1. Imagine exploring the city by moving from one intersection to another.
  2. When you arrive at an intersection, keep track of the fastest time you've reached it and also the second fastest time.
  3. If you reach an intersection faster than your current fastest time, update both the fastest and second fastest times accordingly.
  4. If you reach an intersection, and it's slower than the fastest time, but faster than your second fastest time, update the second fastest time only.
  5. Continue exploring from each intersection, always prioritizing moving in ways that seem fastest.
  6. Stop when you have explored enough that you are confident you've found both the actual fastest and the second fastest times to your destination.
  7. The second fastest time recorded at the destination is the answer.

Code Implementation

from collections import deque

def second_minimum_time_to_reach_destination(number_of_nodes, edges, time, change):

    graph = [[] for _ in range(number_of_nodes + 1)]
    for source, destination in edges:
        graph[source].append(destination)
        graph[destination].append(source)

    shortest_times = [[float('inf')] * 2 for _ in range(number_of_nodes + 1)]
    shortest_times[1][0] = 0

    queue = deque([(1, 0)])

    while queue:
        current_node, current_time = queue.popleft()

        for neighbor in graph[current_node]:
            travel_time = current_time + time

            #Need to wait for green light if red.
            wait_time = (travel_time // change) % 2
            if wait_time == 1:
                travel_time += change - (travel_time % change)

            if travel_time < shortest_times[neighbor][0]:
                shortest_times[neighbor][1] = shortest_times[neighbor][0]
                shortest_times[neighbor][0] = travel_time
                queue.append((neighbor, travel_time))
            elif shortest_times[neighbor][0] < travel_time < shortest_times[neighbor][1]:

                #If it's faster than second best, update.
                shortest_times[neighbor][1] = travel_time
                queue.append((neighbor, travel_time))

    return shortest_times[number_of_nodes][1]

Big(O) Analysis

Time Complexity
O(E + N)The algorithm uses a modified Breadth-First Search (BFS) to explore the graph. In the worst case, it visits each edge and node. E represents the number of roads (edges), and N represents the number of intersections (nodes) in the city. Since each road is considered at most twice (once for each direction), and each intersection is visited at most twice to track the shortest and second shortest times, the overall complexity is proportional to the number of edges plus the number of nodes. Therefore, the time complexity is O(E + N).
Space Complexity
O(N)The algorithm keeps track of the fastest and second fastest times to reach each intersection. This implies storing two time values for each of the N intersections, leading to auxiliary data structures that scale linearly with N. In essence, we are using storage proportional to the number of intersections to remember the best and second-best arrival times. Therefore, the auxiliary space complexity is O(N).

Edge Cases

CaseHow to Handle
Empty graph (n=0)Return -1 immediately as there is no path.
Start and destination are the same node (n=1)Return 0 if time is not 0 or based on constraints if time is specified to be valid.
No path exists between start and destination.Return -1 after BFS exploration if the destination isn't reached.
Graph contains cycles.BFS can handle cycles due to the visited array which prevents revisiting.
Large graph (n is very large) leading to potential memory issues with distances array.Optimize memory usage if needed by storing distance information only for nodes close to the optimal path using bidirectional search if appropriate.
Extreme traffic change value that results in very long wait times.Ensure that the calculations don't cause integer overflow; use long if necessary.
Traffic light change time (change) is zero.Handle this edge case appropriately, likely by returning an error or adjusting the logic to avoid division by zero.
Graph is disconnected.Return -1 after BFS exploration if the destination isn't reached, indicating no path.