#1129

Medium

auto_awesomeGraph traversal with breadth-first search

LeetCode Problem Workspace

Shortest Path with Alternating Colors

Solve the shortest path problem with alternating edge colors using graph traversal and BFS.

Problem Statement

You are given a directed graph with n nodes labeled from 0 to n - 1. The graph has two types of edges: red and blue. You are provided two arrays, redEdges and blueEdges, where each edge connects two nodes. The task is to find the shortest path from node 0 to all other nodes in the graph such that the edge colors alternate along the path, or return -1 if no such path exists.

The function should return an array of length n where each element represents the shortest path to the corresponding node. If no valid alternating path exists from node 0 to a node, return -1 for that node.

Examples

Example 1

Input: n = 3, redEdges = [[0,1],[1,2]], blueEdges = []

Output: [0,1,-1]

Example details omitted.

Example 2

Input: n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]

Output: [0,1,-1]

Example details omitted.

Constraints

1 <= n <= 100
0 <= redEdges.length, blueEdges.length <= 400
redEdges[i].length == blueEdges[j].length == 2
0 <= ai, bi, uj, vj < n

Solution Approach

Breadth-First Search (BFS) with State Tracking

This problem can be approached using BFS, where each state is represented as a tuple consisting of the current node and the last edge color used. The BFS will explore nodes in alternating colors, ensuring the path constraints are met while calculating the shortest distance.

Graph Representation and Alternating Colors

You must represent the graph with two adjacency lists: one for red edges and one for blue edges. In BFS, alternate between exploring the red edges and the blue edges, while maintaining state transitions between nodes and colors.

Queue and Distance Arrays

Use a queue to store states during BFS, and two distance arrays—one for red edges and one for blue edges—to track the shortest distances. The BFS will ensure that we process each node with alternating edge colors correctly.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the graph size and the BFS traversal. Since we visit each node and edge at most once, the time complexity is O(n + E), where n is the number of nodes and E is the total number of edges (red and blue). The space complexity is O(n + E), due to the storage required for the graph representation, distance arrays, and BFS queue.

What Interviewers Usually Probe

Candidates should demonstrate understanding of BFS and how it applies to graph traversal with color alternation.
Look for clarity in representing the problem as a graph and alternating edge colors during BFS.
Candidates should be able to efficiently manage the BFS state transitions between red and blue edges.

Common Pitfalls or Variants

Common pitfalls

Not alternating between red and blue edges properly, causing incorrect paths or infinite loops.
Not handling the case where no valid path exists, leading to incorrect answers.
Mismanaging the state transitions between nodes and edge colors during BFS, causing missed nodes or incorrect distances.

Follow-up variants

Change the problem to find paths with a specific number of alternating edges.
Modify the graph to include weighted edges and ask for the shortest path with alternating edge colors.
Allow multiple starting nodes and find the shortest alternating path from each starting point.

FAQ

What is the best algorithm for solving the Shortest Path with Alternating Colors problem?

The best approach is to use Breadth-First Search (BFS), where each state represents a node and the last edge color used. This ensures the path alternates between red and blue edges.

How can I optimize the space complexity of the solution?

You can optimize space usage by carefully managing the adjacency lists and using a single queue to store BFS states while maintaining only two distance arrays for each edge color.

What are common mistakes when implementing BFS for alternating edge colors?

Common mistakes include not alternating between red and blue edges properly, mishandling edge cases like no valid paths, and incorrectly transitioning between node states in BFS.

Can this problem be solved using Depth-First Search (DFS)?

DFS is not ideal for this problem because it may not efficiently handle the alternating edge constraint. BFS is better suited for finding the shortest path with alternating colors.

What is the time complexity of solving this problem?

The time complexity is O(n + E), where n is the number of nodes and E is the total number of edges (red and blue). This is due to the BFS traversal that processes each node and edge at most once.

terminal

Solution

Solution 1: BFS

The problem is essentially a shortest path problem, which we can consider solving using BFS.

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class Solution:
    def shortestAlternatingPaths(
        self, n: int, redEdges: List[List[int]], blueEdges: List[List[int]]
    ) -> List[int]:
        g = [defaultdict(list), defaultdict(list)]
        for i, j in redEdges:
            g[0][i].append(j)
        for i, j in blueEdges:
            g[1][i].append(j)
        ans = [-1] * n
        vis = set()
        q = deque([(0, 0), (0, 1)])
        d = 0
        while q:
            for _ in range(len(q)):
                i, c = q.popleft()
                if ans[i] == -1:
                    ans[i] = d
                vis.add((i, c))
                c ^= 1
                for j in g[c][i]:
                    if (j, c) not in vis:
                        q.append((j, c))
            d += 1
        return ans

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