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.
[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:
1
and n
.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:
Hence the minimum time needed is 6 minutes.
The red path shows the path to get the second minimum time.
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.
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:
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:
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
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:
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]
Case | How 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. |