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Shortest Path with Alternating Colors

Medium

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Topics:

GraphsArrays

You are given an integer n, the number of nodes in a directed graph where the nodes are labeled from 0 to n - 1. Each edge is red or blue in this graph, and there could be self-edges and parallel edges.

You are given two arrays redEdges and blueEdges where:

redEdges[i] = [a_i, b_i] indicates that there is a directed red edge from node a_i to node b_i in the graph, and
blueEdges[j] = [u_j, v_j] indicates that there is a directed blue edge from node u_j to node v_j in the graph.

Return an array answer of length n, where each answer[x] is the length of the shortest path from node 0 to node x such that the edge colors alternate along the path, or -1 if such a path does not exist.

Example 1:

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

Example 2:

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

Constraints:

1 <= n <= 100
0 <= redEdges.length, blueEdges.length <= 400
redEdges[i].length == blueEdges[j].length == 2
0 <= a_i, b_i, u_j, v_j < n

Try coding it in LeetCode

Shortest Path with Alternating Colors

Medium

Asked by:

27 views

Topics:

GraphsArrays

You are given two arrays redEdges and blueEdges where:

redEdges[i] = [a_i, b_i] indicates that there is a directed red edge from node a_i to node b_i in the graph, and
blueEdges[j] = [u_j, v_j] indicates that there is a directed blue edge from node u_j to node v_j in the graph.

Example 1:

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

Example 2:

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

Constraints:

1 <= n <= 100
0 <= redEdges.length, blueEdges.length <= 400
redEdges[i].length == blueEdges[j].length == 2
0 <= a_i, b_i, u_j, v_j < n

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for 'n' (number of nodes)? Are there any limitations on the number of edges in either the redEdges or blueEdges arrays?
If there is no path with alternating colors between a source node and a destination node, what value should be returned for the distance?
Can there be self-loops in the redEdges or blueEdges arrays? In other words, can an edge connect a node to itself?
Are the graphs directed? (I.e., if (u,v) is in redEdges, does that mean we can only traverse from u to v, or also from v to u using the red edge?)
If multiple shortest paths exist from node 0 to a given node with alternating colors, is it acceptable to return the length of any one of those shortest paths?

Brute Force Solution

Approach

The brute force method for finding the shortest path with alternating colors involves exploring every single possible path through the graph. We will check the color sequence of each path to ensure it alternates, and then identify the shortest of the valid paths found.

Here's how the algorithm would work step-by-step:

Start from the beginning node.
Explore all possible paths emanating from this beginning node, one step at a time.
Each time you move to a new node, remember the color of the edge you just used.
Make sure the next edge you take is a different color than the previous edge.
If the next edge is the same color, then that path is invalid, and you should abandon it.
Continue extending valid paths until you reach all the other nodes.
Record the length of each valid path from the starting node to every other node.
Finally, for each node, select the shortest path length among all valid paths to that node. If no valid path exists to a node, the shortest path length to it is 'not found'.

Code Implementation

def shortest_alternating_paths_brute_force(number_of_nodes, red_edges, blue_edges):
    adjacency_list_red = [[] for _ in range(number_of_nodes)]
    adjacency_list_blue = [[] for _ in range(number_of_nodes)]

    for source, destination in red_edges:
        adjacency_list_red[source].append(destination)
    for source, destination in blue_edges:
        adjacency_list_blue[source].append(destination)

    shortest_paths = [-1] * number_of_nodes
    shortest_paths[0] = 0

    queue = [([0], None)] # (path, last_color) 
    
    while queue:
        current_path, previous_color = queue.pop(0)
        current_node = current_path[-1]

        # Explore red edges if the last edge wasn't red or it's the starting node
        if previous_color != 'red':
            for neighbor in adjacency_list_red[current_node]:
                new_path = current_path + [neighbor]

                if shortest_paths[neighbor] == -1:
                    shortest_paths[neighbor] = len(new_path) - 1

                queue.append((new_path, 'red'))

        # Explore blue edges if the last edge wasn't blue or it's the starting node
        if previous_color != 'blue':
            for neighbor in adjacency_list_blue[current_node]:
                new_path = current_path + [neighbor]

                if shortest_paths[neighbor] == -1:
                    shortest_paths[neighbor] = len(new_path) - 1

                queue.append((new_path, 'blue'))

    return shortest_paths

Big(O) Analysis

Time Complexity

O(2^(|redEdges| + |blueEdges|)) – The brute force approach explores all possible paths in the graph. In the worst case, we might need to examine every combination of red and blue edges. Each edge represents a potential step in a path. The number of possible paths grows exponentially with the total number of edges (|redEdges| + |blueEdges|), as each edge can either be included or excluded in a path. Therefore, the time complexity is O(2^(|redEdges| + |blueEdges|)). This exponential growth makes the brute force method highly inefficient for larger graphs.

Space Complexity

O(N^N) – The brute force approach explores all possible paths from the starting node. In the worst-case scenario, it might need to store a significant number of paths in memory as it explores them. Since each node may have at most N-1 edges and the path length can be at most N in a graph with N nodes, the space used to keep track of all possible paths could grow up to O(N^N), especially if a large branching factor exists. This is because for each node on the path, we're storing all potential outgoing edges and their respective colors. Additionally, the 'shortest path length' for each node needs to be tracked adding a space complexity of O(N).

Optimal Solution

Approach

The best way to find the shortest path with alternating colors is to explore possible paths level by level, remembering the last color used to reach each place. This prevents endless loops and ensures we find the shortest route by only considering paths that alternate colors.

Here's how the algorithm would work step-by-step:

Start at the beginning point, which is the first location.
Imagine exploring the area around each location in a structured way, like ripples spreading in a pond.
When exploring from a location, look at all the possible next steps you can take using one color, say red, then consider those possible steps using the other color, say blue.
Keep track of the last color you used to arrive at each location. This is crucial to avoid going back and forth between the same two locations endlessly.
Only consider moving to a new location if the color you are about to use is different from the color you used to arrive at the current location.
As you explore, record the distance it took to reach each location using an alternating path. If you find a shorter path to a place you've already visited, update the recorded distance.
Continue this process of exploring outward until you have considered all possible paths from the starting point that alternate colors.
The shortest path to any location will be the shortest alternating-color path you found during this exploration.

Code Implementation

def shortest_alternating_path(
    number_of_nodes,
    red_edges,
    blue_edges
):
    adjacency_list_red = [[] for _ in range(number_of_nodes)]
    adjacency_list_blue = [[] for _ in range(number_of_nodes)]

    for source, destination in red_edges:
        adjacency_list_red[source].append(destination)

    for source, destination in blue_edges:
        adjacency_list_blue[source].append(destination)

    shortest_paths = [-1] * number_of_nodes
    shortest_paths[0] = 0

    queue = [(0, None, 0)]
    visited = set()

    while queue:
        node, previous_color, distance = queue.pop(0)

        # Explore red edges if the previous color was not red.
        if previous_color != 'red':
            for neighbor in adjacency_list_red[node]:
                if (neighbor, 'red') not in visited:
                    if shortest_paths[neighbor] == -1:
                        shortest_paths[neighbor] = distance + 1
                    visited.add((neighbor, 'red'))
                    queue.append((neighbor, 'red', distance + 1))

        # Explore blue edges if the previous color was not blue.
        if previous_color != 'blue':
            for neighbor in adjacency_list_blue[node]:
                if (neighbor, 'blue') not in visited:
                    if shortest_paths[neighbor] == -1:
                        shortest_paths[neighbor] = distance + 1
                    visited.add((neighbor, 'blue'))
                    queue.append((neighbor, 'blue', distance + 1))

    return shortest_paths

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm uses a breadth-first search (BFS) approach, exploring the graph level by level. In the worst case, we visit each node (representing a location) and each edge (representing a colored path) at most once. Let 'n' be the number of nodes (locations) and 'm' be the number of edges (red and blue paths combined). The BFS traversal visits each node and edge, leading to a time complexity proportional to the sum of nodes and edges. Thus, the overall time complexity is O(n + m).

Space Complexity

O(N) – The algorithm uses a queue for the breadth-first search, which in the worst-case scenario, could contain all nodes (locations) in the graph. It also uses a data structure (implicitly, a set or matrix) to keep track of visited nodes along with the color used to reach them, preventing cycles. This data structure can store information for each node with each color, leading to a space complexity proportional to the number of nodes, N, and potentially the number of colors (which is constant 2 here). Therefore, the auxiliary space is O(N).

Edge Cases

Case	How to Handle
Null or empty redEdges and blueEdges arrays	Treat null/empty edge lists as graphs with no edges and return appropriate distances.
Graph with only one node (n=1)	Initialize distances array to 0 for the starting node and -1 for others.
Input 'n' is zero or negative	Return an error, throw an exception, or return an empty array depending on the problem's expected behavior.
Cycles exist in the red or blue edges	BFS with color alternation prevents infinite loops by tracking visited nodes and their colors.
No path exists from node 0 to some node i with alternating colors	The distance array will store -1 for those unreachable nodes, indicating no path found.
Integer overflow in distance calculations	The problem statement does not involve large computations, so overflows are not possible.
Disconnected graph, i.e., not all nodes are reachable from node 0.	Unreachable nodes will have distance of -1 in the final result after BFS.
Red and Blue edges connect the same nodes in both directions	BFS will explore both options as they are distinct paths, leading to correct shortest path.

Null or empty redEdges and blueEdges arrays

How to Handle:

Treat null/empty edge lists as graphs with no edges and return appropriate distances.

Graph with only one node (n=1)

How to Handle:

Initialize distances array to 0 for the starting node and -1 for others.

Input 'n' is zero or negative

How to Handle:

Return an error, throw an exception, or return an empty array depending on the problem's expected behavior.

Cycles exist in the red or blue edges

How to Handle:

BFS with color alternation prevents infinite loops by tracking visited nodes and their colors.

No path exists from node 0 to some node i with alternating colors

How to Handle:

The distance array will store -1 for those unreachable nodes, indicating no path found.

Integer overflow in distance calculations

How to Handle:

The problem statement does not involve large computations, so overflows are not possible.

Disconnected graph, i.e., not all nodes are reachable from node 0.

How to Handle:

Unreachable nodes will have distance of -1 in the final result after BFS.

Red and Blue edges connect the same nodes in both directions

How to Handle:

BFS will explore both options as they are distinct paths, leading to correct shortest path.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for 'n' (number of nodes)? Are there any limitations on the number of edges in either the redEdges or blueEdges arrays?
If there is no path with alternating colors between a source node and a destination node, what value should be returned for the distance?
Can there be self-loops in the redEdges or blueEdges arrays? In other words, can an edge connect a node to itself?
Are the graphs directed? (I.e., if (u,v) is in redEdges, does that mean we can only traverse from u to v, or also from v to u using the red edge?)
If multiple shortest paths exist from node 0 to a given node with alternating colors, is it acceptable to return the length of any one of those shortest paths?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start from the beginning node.
Explore all possible paths emanating from this beginning node, one step at a time.
Each time you move to a new node, remember the color of the edge you just used.
Make sure the next edge you take is a different color than the previous edge.
If the next edge is the same color, then that path is invalid, and you should abandon it.
Continue extending valid paths until you reach all the other nodes.
Record the length of each valid path from the starting node to every other node.
Finally, for each node, select the shortest path length among all valid paths to that node. If no valid path exists to a node, the shortest path length to it is 'not found'.

Code Implementation

def shortest_alternating_paths_brute_force(number_of_nodes, red_edges, blue_edges):
    adjacency_list_red = [[] for _ in range(number_of_nodes)]
    adjacency_list_blue = [[] for _ in range(number_of_nodes)]

    for source, destination in red_edges:
        adjacency_list_red[source].append(destination)
    for source, destination in blue_edges:
        adjacency_list_blue[source].append(destination)

    shortest_paths = [-1] * number_of_nodes
    shortest_paths[0] = 0

    queue = [([0], None)] # (path, last_color) 
    
    while queue:
        current_path, previous_color = queue.pop(0)
        current_node = current_path[-1]

        # Explore red edges if the last edge wasn't red or it's the starting node
        if previous_color != 'red':
            for neighbor in adjacency_list_red[current_node]:
                new_path = current_path + [neighbor]

                if shortest_paths[neighbor] == -1:
                    shortest_paths[neighbor] = len(new_path) - 1

                queue.append((new_path, 'red'))

        # Explore blue edges if the last edge wasn't blue or it's the starting node
        if previous_color != 'blue':
            for neighbor in adjacency_list_blue[current_node]:
                new_path = current_path + [neighbor]

                if shortest_paths[neighbor] == -1:
                    shortest_paths[neighbor] = len(new_path) - 1

                queue.append((new_path, 'blue'))

    return shortest_paths

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning point, which is the first location.
Imagine exploring the area around each location in a structured way, like ripples spreading in a pond.
When exploring from a location, look at all the possible next steps you can take using one color, say red, then consider those possible steps using the other color, say blue.
Keep track of the last color you used to arrive at each location. This is crucial to avoid going back and forth between the same two locations endlessly.
Only consider moving to a new location if the color you are about to use is different from the color you used to arrive at the current location.
As you explore, record the distance it took to reach each location using an alternating path. If you find a shorter path to a place you've already visited, update the recorded distance.
Continue this process of exploring outward until you have considered all possible paths from the starting point that alternate colors.
The shortest path to any location will be the shortest alternating-color path you found during this exploration.

Code Implementation

def shortest_alternating_path(
    number_of_nodes,
    red_edges,
    blue_edges
):
    adjacency_list_red = [[] for _ in range(number_of_nodes)]
    adjacency_list_blue = [[] for _ in range(number_of_nodes)]

    for source, destination in red_edges:
        adjacency_list_red[source].append(destination)

    for source, destination in blue_edges:
        adjacency_list_blue[source].append(destination)

    shortest_paths = [-1] * number_of_nodes
    shortest_paths[0] = 0

    queue = [(0, None, 0)]
    visited = set()

    while queue:
        node, previous_color, distance = queue.pop(0)

        # Explore red edges if the previous color was not red.
        if previous_color != 'red':
            for neighbor in adjacency_list_red[node]:
                if (neighbor, 'red') not in visited:
                    if shortest_paths[neighbor] == -1:
                        shortest_paths[neighbor] = distance + 1
                    visited.add((neighbor, 'red'))
                    queue.append((neighbor, 'red', distance + 1))

        # Explore blue edges if the previous color was not blue.
        if previous_color != 'blue':
            for neighbor in adjacency_list_blue[node]:
                if (neighbor, 'blue') not in visited:
                    if shortest_paths[neighbor] == -1:
                        shortest_paths[neighbor] = distance + 1
                    visited.add((neighbor, 'blue'))
                    queue.append((neighbor, 'blue', distance + 1))

    return shortest_paths

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty redEdges and blueEdges arrays	Treat null/empty edge lists as graphs with no edges and return appropriate distances.
Graph with only one node (n=1)	Initialize distances array to 0 for the starting node and -1 for others.
Input 'n' is zero or negative	Return an error, throw an exception, or return an empty array depending on the problem's expected behavior.
Cycles exist in the red or blue edges	BFS with color alternation prevents infinite loops by tracking visited nodes and their colors.
No path exists from node 0 to some node i with alternating colors	The distance array will store -1 for those unreachable nodes, indicating no path found.
Integer overflow in distance calculations	The problem statement does not involve large computations, so overflows are not possible.
Disconnected graph, i.e., not all nodes are reachable from node 0.	Unreachable nodes will have distance of -1 in the final result after BFS.
Red and Blue edges connect the same nodes in both directions	BFS will explore both options as they are distinct paths, leading to correct shortest path.

Null or empty redEdges and blueEdges arrays

How to Handle:

Treat null/empty edge lists as graphs with no edges and return appropriate distances.

Graph with only one node (n=1)

How to Handle:

Initialize distances array to 0 for the starting node and -1 for others.

Input 'n' is zero or negative

How to Handle:

Return an error, throw an exception, or return an empty array depending on the problem's expected behavior.

Cycles exist in the red or blue edges

How to Handle:

BFS with color alternation prevents infinite loops by tracking visited nodes and their colors.

No path exists from node 0 to some node i with alternating colors

How to Handle:

The distance array will store -1 for those unreachable nodes, indicating no path found.

Integer overflow in distance calculations

How to Handle:

The problem statement does not involve large computations, so overflows are not possible.

Disconnected graph, i.e., not all nodes are reachable from node 0.

How to Handle:

Unreachable nodes will have distance of -1 in the final result after BFS.

Red and Blue edges connect the same nodes in both directions

How to Handle:

BFS will explore both options as they are distinct paths, leading to correct shortest path.