How do you solve Shortest Path with Alternating Colors on LeetCode?

Shortest Path with Alternating Colors (LeetCode #1129) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + e) time complexity and O(n) space complexity. Key insight: Use BFS for shortest path problems in graphs.

What is the time complexity of Shortest Path with Alternating Colors?

The optimal solution for Shortest Path with Alternating Colors runs in O(n + e) time and uses O(n) space. The complexity is O(n + e) where e is the number of edges, as we explore each node and edge once during BFS.

What is the brute force solution for Shortest Path with Alternating Colors?

The brute force approach for Shortest Path with Alternating Colors is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can explore all possible paths starting from node 0. We will keep track of the colors of the edges taken and ensure they alternate. This approach is simple but inefficient as it may explore many unnecessary paths. The optimal Optimal Solution approach improves this to O(n + e).

What algorithm or data structure is used to solve Shortest Path with Alternating Colors?

Shortest Path with Alternating Colors uses the following concepts: Breadth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Breadth-First Search, State Management in Graphs.

What are the interview tips for Shortest Path with Alternating Colors?

Explain your thought process clearly while coding. Consider edge cases, such as no edges or self-loops. Practice BFS and graph traversal problems to build intuition.

What are common mistakes when solving Shortest Path with Alternating Colors?

Not alternating colors correctly. Forgetting to check if a node has been visited.

#1129

Shortest Path with Alternating Colors

Medium

Breadth-First SearchGraph TheoryGraph TraversalBreadth-First SearchState Management in Graphs

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + e)
Space	O(1)	O(n)

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Intuition

Time O(n + e)Space O(n)

The optimal solution uses a breadth-first search (BFS) approach, where we maintain two queues to track the current node and the last edge color used. This allows us to efficiently explore the graph while ensuring that colors alternate, leading to a faster solution.

⚙️

Algorithm

3 steps

1Step 1: Create a graph representation using adjacency lists for both red and blue edges.
2Step 2: Initialize a queue for BFS, starting from node 0 with both colors and a distance of 0.
3Step 3: While the queue is not empty, explore neighbors based on the last edge color, updating distances and adding new nodes to the queue.

solution.py23 lines

1# Full working Python code
2from collections import deque
3
4def shortestAlternatingPaths(n, redEdges, blueEdges):
5    graph = {i: {'red': [], 'blue': []} for i in range(n)}
6    for u, v in redEdges:
7        graph[u]['red'].append(v)
8    for u, v in blueEdges:
9        graph[u]['blue'].append(v)
10
11    distances = [-1] * n
12    distances[0] = 0
13    queue = deque([(0, 'red', 0), (0, 'blue', 0)])
14
15    while queue:
16        node, color, dist = queue.popleft()
17        next_color = 'blue' if color == 'red' else 'red'
18        for neighbor in graph[node][next_color]:
19            if distances[neighbor] == -1:
20                distances[neighbor] = dist + 1
21                queue.append((neighbor, next_color, dist + 1))
22
23    return distances

ℹ

Complexity note: The complexity is O(n + e) where e is the number of edges, as we explore each node and edge once during BFS.

1Use BFS for shortest path problems in graphs.
2Maintain state of the last edge color to ensure alternation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.