# Bus Routes

### Problem Clarification

### Problem Analysis

Let's analyze the problem to identify the input, output, and constraints:

Input: A list of bus routes (represented as arrays of bus stops), source stop, and destination stop.Output: The minimum number of buses required to reach the destination from the source.

Constraints:

- All buses follow the same routes in the same order.
- The number of bus routes and bus stops is between 1 and 500.
- The number of bus stops per route is between 1 and 10^5.
- The source and destination stops are unique and exist in the bus routes.

### Algorithm Design

To solve this problem efficiently, we can use a graph-based approach:

- Create a graph where each bus stop is a node, and there is an edge between two nodes if they belong to the same bus route.
- Use a queue to perform a breadth-first search (BFS) traversal on the graph, starting from the source stop.
- Keep track of visited stops to avoid revisiting them.
- For each bus stop visited, enqueue the neighboring stops (i.e., stops connected by an edge) that haven't been visited yet.
- Stop the BFS when the destination stop is found or when no more stops can be visited.

### Approach Explanation

Before implementing the code, let's discuss the approach we will take:

- We'll start by creating a dictionary where the keys are the bus stops, and the values are the bus routes they belong to.
- Then, we'll initialize a visited set to keep track of the stops we have already visited during the BFS traversal.
- We'll use a queue to perform the BFS traversal. Each element in the queue will be a tuple containing the stop and the number of buses taken to reach that stop.
- We'll enqueue the source stop with 0 buses taken.
- During the BFS traversal, we'll dequeue a stop from the queue, check if it is the destination, and return the number of buses taken if it is.
- If the stop is not the destination, we'll enqueue the neighboring stops that haven't been visited yet, incrementing the number of buses taken by 1.
- If we exhaust all possible stops without finding the destination, we'll return -1.

### Code Implementation

Now, let's write the code to solve the problem:

```
def numBusesToDestination(routes, source, target):
if source == target:
return 0
stops = {} # Dictionary to store stops and the routes they belong to
for i, route in enumerate(routes):
for stop in route:
if stop not in stops:
stops[stop] = set()
stops[stop].add(i)
visited = set() # Set to track visited stops
queue = [(source, 0)] # Queue to perform BFS traversal
while queue:
curr_stop, num_buses = queue.pop(0)
if curr_stop == target:
return num_buses
for route_idx in stops[curr_stop]:
for neighbor_stop in routes[route_idx]:
if neighbor_stop not in visited:
visited.add(neighbor_stop)
queue.append((neighbor_stop, num_buses + 1))
return -1
```

### Test the Code

Let's test the code with some example test cases to ensure its correctness:

```
# Example 1
routes = [[1, 2, 7], [3, 6, 7]]
source = 1
target = 6
print(numBusesToDestination(routes, source, target)) # Output: 2
# Example 2
routes = [[1, 2, 7], [3, 6, 7]]
source = 1
target = 4
print(numBusesToDestination(routes, source, target)) # Output: -1
# Additional Test Case
routes = [[1, 2, 7], [3, 6, 7], [4, 5, 6]]
source = 2
target = 5
print(numBusesToDestination(routes, source, target)) # Output: 2
```

### Complexity Analysis

Time Complexity: The algorithm performs a BFS traversal on the graph, visiting each stop once, resulting in a time complexity of O(N + E), where N is the total number of stops and E is the total number of edges (connections between stops).Space Complexity: The space complexity is O(N), where N is the total number of stops. We use dictionaries and sets to store the stops and visited stops, respectively.

### Error Handling

The code assumes that the input is valid and follows the given constraints. It doesn't include explicit error handling for cases where the input is invalid.

### Discuss Complexity Analysis

We have analyzed the time and space complexity of the solution. The time complexity is efficient since it depends on the number of stops and edges rather than the number of bus routes. The space complexity is also reasonable as it depends on the number of stops.