# Solution For Shortest Path With Alternating Colors

Problem Statement

The problem asks us to find the shortest path between two nodes in a graph where edges are of two colors, red and blue. We can only traverse edges of a specific color at a time, i.e., if we traverse a red edge, we can only traverse the next edge of blue color, and vice versa.

Approach

To solve the problem, we can use Breadth First Search (BFS) algorithm where we would keep track of the current color while traversing the graph. For this, we can maintain two queues, one for blue color edges, and the other for red color edges. We would also maintain two distance arrays, one for the blue color edges, and the other for red color edges.

We would start by traversing the graph from the given start node and initialize the distance arrays for both colors. We would then push the start node to both queues and set its distance as zero.

As we traverse the graph using BFS, we would keep track of the current color. If we traverse a node using a red color edge, we would push its adjacent nodes to the blue queue with distance as the current distance + 1 (since we have to traverse a blue edge after we traverse the current red edge). Similarly, if we traverse a node using a blue color edge, we would push its adjacent nodes to the red queue with distance as the current distance + 1.

We would continue the BFS until we reach the target node or both queues become empty (which implies that we have traversed the entire graph and the target node is not reachable).

If we reach the target node, we would return the minimum distance from the both the distance arrays (as we can reach the target node using either red or blue edges). If the target node is not reachable, we would return -1.

Code Implementation

We would start by creating a graph using adjacency list representation.

graph = [[] for _ in range(n)]

for u, v, color in red_edges + blue_edges:

if color == 1:

graph[u].append((v, True)) # adding red color edge

else:

graph[u].append((v, False)) # adding blue color edge

We would then define two queues, two distance arrays, and a set to keep track of visited nodes.

red_queue = collections.deque([(start, 0)])

blue_queue = collections.deque([(start, 0)])

red_dist = [float(‘inf’)] * n

blue_dist = [float(‘inf’)] * n

visited = set()

We would then start the BFS from the start node.

while red_queue or blue_queue:

# process red queue

while red_queue:

node, dist = red_queue.popleft()

if node not in visited:

visited.add(node)

red_dist[node] = min(red_dist[node], dist)

for adj_node, color in graph[node]:

if not color:

blue_queue.append((adj_node, dist+1))

```
# process blue queue
while blue_queue:
node, dist = blue_queue.popleft()
if node not in visited:
visited.add(node)
blue_dist[node] = min(blue_dist[node], dist)
for adj_node, color in graph[node]:
if color:
red_queue.append((adj_node, dist+1))
```

As we traverse the graph using BFS, we keep pushing nodes to the respective queues and updating the distance arrays. We also keep track of the visited nodes.

Once we have finished the BFS traversal, we would check if the target node is reachable. If it is reachable, we would return the minimum distance from the distance arrays. Otherwise, we would return -1.

if red_dist[dest] != float(‘inf’) or blue_dist[dest] != float(‘inf’):

return min(red_dist[dest], blue_dist[dest])

else:

return -1

Complete Code

Here is the complete code for the problem:

import collections

class Solution:

def shortestAlternatingPaths(self, n: int, red_edges: List[List[int]], blue_edges: List[List[int]]) -> List[int]:

```
# create graph
graph = [[] for _ in range(n)]
for u, v, color in red_edges + blue_edges:
if color == 1:
graph[u].append((v, True)) # adding red color edge
else:
graph[u].append((v, False)) # adding blue color edge
# initialize queues and distance arrays
red_queue = collections.deque([(0, 0)])
blue_queue = collections.deque([(0, 0)])
red_dist = [float('inf')] * n
blue_dist = [float('inf')] * n
visited = set()
# start BFS traversal
while red_queue or blue_queue:
# process red queue
while red_queue:
node, dist = red_queue.popleft()
if node not in visited:
visited.add(node)
red_dist[node] = min(red_dist[node], dist)
for adj_node, color in graph[node]:
if not color:
blue_queue.append((adj_node, dist+1))
# process blue queue
while blue_queue:
node, dist = blue_queue.popleft()
if node not in visited:
visited.add(node)
blue_dist[node] = min(blue_dist[node], dist)
for adj_node, color in graph[node]:
if color:
red_queue.append((adj_node, dist+1))
# check if target node is reachable
if red_dist[n-1] != float('inf') or blue_dist[n-1] != float('inf'):
return min(red_dist[n-1], blue_dist[n-1])
else:
return -1
```

Time Complexity

The time complexity of the solution is O(E + V), where E is the number of edges and V is the number of vertices in the graph. This is because we are traversing each edge and vertex once using Breadth First Search algorithm.

Space Complexity

The space complexity of the solution is O(E + V), which is the maximum amount of space used by the visited set, distance arrays, and queues.

## Step by Step Implementation For Shortest Path With Alternating Colors

There are two different shortest path algorithms that could be used for this problem, Dijkstra's algorithm and Bellman-Ford algorithm. However, since the graph is unweighted, Dijkstra's algorithm would be the better choice. The basic idea behind Dijkstra's algorithm is to find the shortest path from a single source vertex to all other vertices in the graph. We can use a min-heap to keep track of the vertices that have been visited, and the order in which they were visited. The min-heap will allow us to always visit the vertex that is closest to the source vertex. We can also keep track of the colors of the vertices in a separate array. We start by adding the source vertex to the min-heap, and setting the color of the source vertex to 0. Then, we do the following until the min-heap is empty: 1. Remove the vertex v from the min-heap. 2. For each neighbor w of v: If the color of w is different from the color of v: Set the color of w to the color of v. Update the distance of w (in the min-heap) to be the distance of v plus 1. 3. Set the color of v to -1 (indicating that it has been visited). The time complexity of this algorithm is O(E + VlogV), where E is the number of edges and V is the number of vertices.

There are a few things to consider when solving this problem. First, we need to keep track of which nodes have been visited in order to avoid cycles. Second, we need to keep track of the shortest path to each node. Third, we need to be able to alternate colors as we traverse the graph. We can use a breadth-first search to find the shortest path from the starting node to each other node. To keep track of which nodes have been visited, we can use a set. To keep track of the shortest path to each node, we can use a dictionary. def shortest_path_with_alternating_colors(graph, start): # keep track of which nodes have been visited visited = set() # keep track of the shortest path to each node shortest_path = {} # enqueue the starting node queue = [start] # set the distance to the starting node to 0 shortest_path[start] = 0 # set the color to 0 (red) color = 0 # while the queue is not empty while queue: # dequeue a node node = queue.pop(0) # if the node has not been visited if node not in visited: # mark the node as visited visited.add(node) # for each neighbor of the node for neighbor in graph[node]: # if the neighbor has not been visited if neighbor not in visited: # enqueue the neighbor queue.append(neighbor) # set the distance to the neighbor shortest_path[neighbor] = shortest_path[node] + 1 # set the color to the opposite of the current color color = 1 - color # return the shortest path return shortest_path

There are two possible colors that nodes can have in this problem, red and blue. We can keep track of the shortest path to each node by using a 2D array, where each element in the array represents the shortest path to that node using that color. We can start by initializing the 2D array with Infinity for all values. Then, we can use a breadth-first search algorithm to find the shortest path to each node. We can keep track of which nodes have been visited in a separate array. As we visit each node, we will check the value of the shortest path for that node using the other color. If the value is Infinity, that means we have not yet visited that node using that color. We will update the value in the 2D array and add the node to our queue. Once we have visited all nodes, we can return the value of the shortest path for the last node in the 2D array.

There are two types of nodes in this graph, red nodes and blue nodes. The graph is given as follows: the nodes are labeled with integers from 1 to n. Red nodes are labeled with odd integers, blue nodes are labeled with even integers. So, for example, if n=5, then the red nodes are labeled 1, 3, and 5, and the blue nodes are labeled 2 and 4. Given a graph with n nodes and m edges, find the shortest path from node 1 to node n that visits red nodes and blue nodes alternately. If such a path does not exist, return -1. The solution to this problem is as follows: 1) Use BFS to find the shortest path from node 1 to node n. 2) At each node, check if it is red or blue. 3) If it is red, add it to the queue of red nodes. 4) If it is blue, add it to the queue of blue nodes. 5) Alternate between the queues, popping a node from the front of each queue in turn. 6) If the node being popped is the goal node, then we have found the shortest path and can return it. 7) If we reach the end of one of the queues without finding the goal node, then we know that there is no path that alternates between red and blue nodes, and we can return -1.

There are two possible colors that nodes can have - red or blue. Given a graph with n nodes and m edges, each node is initially colored red. You need to find the shortest path from node 0 to node n-1, such that at any point along the path, no two adjacent nodes have the same color. If such a path does not exist, return -1. The graph is represented as an adjacency matrix, where the value at index [i, j] indicates the weight of the edge between nodes i and j (0 if there is no edge). Example 1: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] Output: 200 Explanation: The graph looks like this: The cheapest path from node 0 to node 2 is 0 -> 1 -> 2, and the cost is 100 + 100 = 200. Example 2: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] Output: 500 Explanation: The graph looks like this: The cheapest path from node 0 to node 2 is 0 -> 1 -> 2, and the cost is 100 + 100 = 200. Example 3: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] Output: -1 Explanation: The graph looks like this: The cheapest path from node 0 to node 2 is 0 -> 1 -> 2, and the cost is 100 + 100 = 200.