Problem Description

In this problem, we have a directed graph of n nodes, labeled from 0 to n - 1. There are two types of directed edges in the graph: red edges and blue edges. A red edge connects two nodes in one direction, i.e., redEdges[i] = [a_i, b_i] represents a red edge from node a_i to node b_i. Similarly, blueEdges[j] = [u_j, v_j] represents a blue edge from node u_j to node v_j.

The objective is to find the shortest path from node 0 to every other node such that the colors of the edges alternate along the path. We need to return an array answer of length n, where answer[x] is the length of such shortest path to node x. If there is no path from node 0 to node x with alternating edge colors, we set answer[x] to -1.

The challenge lies in navigating the graph in a way that each edge color along the path alternates and ensuring we are tracking the shortest such path for each node.

Flowchart Walkthrough

Let's determine the appropriate algorithm using the Flowchart. Here’s a detailed analysis:

Is it a graph?

• Yes: The problem models cities connected by red and blue-colored edges in a directed graph.

Is it a tree?

• No: Although the graph is directed, it’s not necessarily acyclic and can have multiple paths between nodes, indicating it's not a tree structure.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem deals with potentially cyclical paths due to the alternating colors of the edges, not just acyclic paths.

Is the problem related to shortest paths?

• Yes: The objective is to find the shortest path from node 0 to all other nodes using edges of alternating colors.

Is the graph weighted?

• No: The problem does not specify any weights on the edges; the concern is purely with the path lengths in terms of the number of edges traversed.

Does the problem involve connectivity?

• The focus is more on finding shortest paths rather than on connectivity or finding connected components. Although BFS can be employed in both shortest paths and connectivity problems, here we are specifically concerned with shortest paths in an unweighted graph.

Conclusion: The flowchart suggests using BFS for this unweighted shortest path problem, specifically targeting a scenario where only the number of edges (steps through alternating colors) counts, and not any additional weighings on those edges.

Intuition

To solve this problem, we leverage Breadth First Search (BFS). BFS is a well-suited algorithm for finding the shortest path in an unweighted graph, which aligns with our goal of finding the shortest alternating paths.

The intuition behind using BFS for this problem stems from its level-by-level traversal, which allows us to keep track of the path length naturally. To account for the alternating colors, we modify the traditional BFS by keeping a separate queue for red and blue edges and toggling between them on each step.

Furthermore, we need to prevent revisiting the same node through the same color edge to avoid potential infinite loops. To do this, we maintain a visited set that records nodes visited through a specific color edge.

In the solution code, we have two different graph representations, g[0] for red edges and g[1] for blue edges. Then, we start BFS from node 0 using two entries, one for a red edge and one for a blue edge, assuming an imaginary edge of both colors leading to 0. We increment the distance d each time we finish a level in the BFS traversal.

The vis set keeps track of nodes we have visited with a specific color, and the q deque contains nodes to be visited with the associated color. We use the c variable to represent the color and toggle it using c ^= 1 (where 0 represents red and 1 represents blue). If we reach a new node, we update the answer, and each new edge found is added to the queue to continue the search.

By the end of the BFS procedure, we obtain an array ans where each ans[i] is the shortest alternating path length to node i or -1 if not reachable.

Learn more about Breadth-First Search and Graph patterns.

Solution Approach

The solution uses a modified Breadth-First Search (BFS) algorithm to ensure path length is minimized and the alternating color requirement is met. Below is an explanation of the various parts of the solution and how they implement this modified BFS strategy:

• Graph Representation: We use two defaultdict(list) to represent the directed graph, one for red edges (g[0]) and one for blue edges (g[1]). This allows us to easily access all the outgoing edges of a particular color from any node.

• Initialization: The ans array is initialized to -1 for each node. This signifies that initially, all nodes are considered unreachable from node 0 via an alternating path. We also initialize a set vis to keep track of visited (node, color) pairs to prevent reprocessing, and a double-ended queue q with two tuples representing the initial node 0 with both red and blue colors.

• BFS Loop:

  • We only increase our distance counter d once we've looked at all nodes in our queue, essential for ensuring that BFS traverses the graph level-by-level and correctly calculates the shortest path lengths.
  • For each node we dequeue (i) and its associated color (c), we check if we have already recorded a path to that node. If not, we record the current distance d as the shortest path length to i.
  • We add the node to the visited set vis to avoid revisiting a node via the same color edge.
  • To enforce alternate coloring, we toggle the color c using c ^= 1, which switches 0 to 1 and 1 to 0.

• Queue Management: We iterate over each outgoing edge of the toggled color from node i. If the adjacent node j through color c hasn't been visited, we add (j, c) to the queue q to visit it in subsequent iterations.

• Result: After the BFS terminates, the ans array contains the shortest path lengths from node 0 to all others with alternating colors, or -1 if no path exists. This array is then returned as the result.

In short, the key to this solution is managing a queue that tracks the next node to explore and the color of the edge we used to reach it, while also toggling the color on each step to ensure the alternation of red and blue edges.

Example Walkthrough

Let's take a small directed graph as an example with 5 nodes (0 to 4) and alternating red and blue edges as follows:

Red edges: redEdges = [[0, 1], [1, 2], [2, 3]] Blue edges: blueEdges = [[1, 0], [2, 1], [3, 4]]

We start searching from node 0 with the objective to find the shortest alternating path from node 0 to all other nodes.

Initialization:

• Start with ans array as [-1, -1, -1, -1, -1], since initially we have not found any paths.
• The queue q is initialized with two tuples: [(0, 0), (0, 1)] indicating node 0 reachable by both red and blue edges.
• The visited set vis is empty.

BFS Loop:

1. Dequeue (0, 0), representing the starting node with a red edge. There is no actual red edge, but conceptually we use this to start the search.
2. We mark node 0 as visited with a red edge by adding (0, 0) to vis.
3. Since 0 is the starting node, ans[0] is set to 0.
4. We look at blue edges from 0 since we need to alternate, but there are none.
5. Dequeue the next element (0, 1), representing the starting node with a blue edge.
6. We mark node 0 as visited with a blue edge by adding (0, 1) to vis.
7. Similar to before, ans[0] is 0.
8. We look for red edges from 0 and find 0 -> 1. Since (1, 0) is not visited, we add (1, 0) to q.
9. Increment d to 1 since we've finished all nodes at the current level.

Next level:

1. Dequeue (1, 0); we note that node 1 hasn't been visited from a red edge, so we add it to vis as (1, 0) and set ans[1] to d, which is 1.
2. We toggle the color to blue and look for blue edges from 1, which is 1 -> 0. However, this leads back to node 0, which has already been visited by a blue edge.

Level d = 2:

1. We keep processing the remaining nodes in the queue similarly, ensuring that we only move to the next level once all nodes in q from the previous level are processed.
2. Eventually, we discover 2 -> 1 and 3 -> 4, so ans[2] is set to 2 and ans[3] to 2 as well. Node 4 is reached by a blue edge from node 3, so ans[4] is set to 3.

The BFS loop continues until all possible nodes have been visited via alternating paths or the queue is empty.

Final ans array after running the BFS will be [0, 1, 2, 2, 3]. This array indicates the shortest paths available from node 0 to all other nodes via alternating red and blue edges, with the corresponding lengths of those paths. If we had not been able to reach a node with alternating paths, that value would remain as -1.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(V + E) where V is the number of vertices (nodes) and E is the number of edges. This is because each node is enqueued and dequeued at most once for each color (since the algorithm keeps track of the color of the paths). The inner loop, which iterates over the queue q, runs for each edge at most once for each color as well because an edge will only be followed if it provides a path of the opposite color from the current node.

Space Complexity

The space complexity is O(V + E) as well. This comes from the storage required for:

• g: adjacency lists for both red and blue edges, each of which contains at most E pairs.
• ans: an array containing n elements.
• vis: a set containing tuples for visited nodes with a particular color; at most 2V tuples will be stored because each node can be visited with at most two colors (red and blue).
• q: a deque that can contain at most 2V elements, since it holds pairs of node indices and color.

In summary, the structures proportional to E in the worst-case scenario dominate the space complexity, and hence it is O(V + E).

Learn more about how to find time and space complexity quickly using problem constraints.