Largest Color Value in a Directed Graph

Hard

Largest Color Value in a Directed Graph

Hard

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 maximum number of nodes and edges in the graph? What is the maximum length of the colors string?
Can the graph contain cycles? If so, what should the function return in the presence of a cycle?
Is the input 'colors' string guaranteed to only contain lowercase English letters?
If there are multiple paths with the same largest color value, can I return any one of them?
If the graph is empty (no nodes or edges), what value should I return?

Brute Force Solution

Approach

The brute force approach in this scenario is like trying every single possible path through the graph to find the one with the highest color value. We explore all routes, regardless of how inefficient or long they might be. Eventually we should find a path that gives us the biggest color value.

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

Start at each node in the graph, one at a time.
From each starting node, explore all possible paths that can be taken by following the graph's connections.
For each path, count the occurrences of each color along that path.
Determine the color that appears most often in that path, and remember how many times it appears.
Once all possible paths from all starting nodes have been explored, compare the maximum color counts obtained from each path.
The largest of these counts is the largest color value in the directed graph.

Code Implementation

def largest_color_value_brute_force(colors, edges):
    number_of_nodes = len(colors)
    adjacency_list = [[] for _ in range(number_of_nodes)]

    for start_node, end_node in edges:
        adjacency_list[start_node].append(end_node)

    maximum_color_value = 0

    for start_node in range(number_of_nodes):
        # Need to check all possible paths starting from each node
        def depth_first_search(current_node, current_path, color_counts):
            nonlocal maximum_color_value

            current_path.append(current_node)
            color_counts[colors[current_node]] += 1

            maximum_color_value = max(maximum_color_value, max(color_counts.values()))

            # Avoid cycles by checking if current_node is in current_path
            for neighbor in adjacency_list[current_node]:
                if neighbor not in current_path:
                    depth_first_search(neighbor, current_path.copy(), color_counts.copy())

        depth_first_search(start_node, [], {color: 0 for color in set(colors)})

    return maximum_color_value

Big(O) Analysis

Time Complexity

O(n * m * 26) – The brute force approach explores all possible paths from each of the n nodes. In the worst case, each node could potentially have a path visiting all other nodes, leading to a cost dependent on the number of edges m and the length of the longest possible path (which can be up to n in a graph with cycles). For each of these paths, we count the occurrences of each color. Since there are 26 possible colors, this color counting takes O(26) which is O(1), but we include it for completeness. Multiplying these factors gives us approximately n * m * 26 operations. This simplifies to O(n * m * 26), and since 26 is a constant, we can simplify to O(n * m). However, given the provided explanation, we should include the factor for the color counting.

Space Complexity

O(N) – The brute force approach explores all possible paths starting from each node. To keep track of the visited nodes in each path and the color counts along that path, auxiliary space is used. In the worst case, a path could visit all N nodes in the graph, requiring space to store the path and the color counts for each node in the path, thus leading to O(N) space. The color counts can be stored in a hash map or array, also of size at most N since there can be at most N unique colors in the path.

Optimal Solution

Approach

This problem wants us to find the largest number of times a color appears on any path through a colored graph. The trick is to process nodes in an order that guarantees we've already computed the best color counts for all incoming nodes before considering a node itself.

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

First, figure out which nodes have no incoming connections. These are our starting points.
Keep track of how many connections each node receives from other nodes. This helps us know when we've processed all incoming nodes.
Start with a node that has no incoming connections. For each color, keep track of the maximum times that color appears on a path ending at that node.
Visit each node connected to the current node.
Update the maximum color counts for the connected node, considering the path from the starting node, making sure to update color counts to be the largest seen thus far.
Decrease the number of incoming connections for the connected node, because we have processed one connection now.
If any node's incoming connection count drops to zero, add it to the list of nodes we can process next.
Repeat the process of selecting nodes with no incoming connections and updating color counts until all nodes have been processed.
If you ever find a cycle in the graph (meaning you can't process all the nodes), it means the color count can increase indefinitely, so report that there is no largest color value.
Finally, find the largest color count across all the nodes. This is the answer.

Code Implementation

def largestColorValue(colors, edges):
    number_of_nodes = len(colors)
    adjacency_list = [[] for _ in range(number_of_nodes)]
    in_degree = [0] * number_of_nodes

    for start_node, end_node in edges:
        adjacency_list[start_node].append(end_node)
        in_degree[end_node] += 1

    queue = [node for node in range(number_of_nodes) if in_degree[node] == 0]
    color_counts = [[0] * 26 for _ in range(number_of_nodes)]

    # Initialize color counts for starting nodes
    for node in queue:
        color_counts[node][ord(colors[node]) - ord('a')] = 1

    visited_nodes = 0
    maximum_color_value = 0

    while queue:
        current_node = queue.pop(0)
        visited_nodes += 1

        maximum_color_value = max(maximum_color_value,
                                max(color_counts[current_node]))

        for neighbor_node in adjacency_list[current_node]:

            # Propagate color counts to neighbors
            for color_index in range(26):
                color_counts[neighbor_node][color_index] = max(
                    color_counts[neighbor_node][color_index],
                    color_counts[current_node][color_index] +
                    (1 if ord(colors[neighbor_node]) - ord('a') == color_index else 0)
                )

            in_degree[neighbor_node] -= 1

            # Enqueue nodes with no more incoming edges
            if in_degree[neighbor_node] == 0:
                queue.append(neighbor_node)

    # Cycle detected if not all nodes were visited
    if visited_nodes != number_of_nodes:
        return -1

    return maximum_color_value

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm involves processing each node and each edge of the graph. We iterate through all n nodes to initialize in-degrees and the adjacency list representing the m edges. The topological sort-like traversal visits each node once and each edge once when updating color counts. The operation of finding the maximum color count at the end takes O(n). Therefore, the overall time complexity is dominated by the traversal of nodes and edges, resulting in O(n + m).

Space Complexity

O(N) – The algorithm utilizes several auxiliary data structures. It maintains a list of nodes with no incoming connections, which can potentially hold all N nodes in the worst case. It also keeps track of incoming connections for each node, requiring space proportional to N. Additionally, for each node, it stores the maximum color counts, leading to another data structure of size N * number of colors, which simplifies to N since the number of colors is constant. Therefore, the dominant space complexity is O(N).

Edge Cases

Case	How to Handle
Empty graph (n = 0, edges is empty)	Return 0, as no nodes or colors exist.
Graph with a single node and a color	Return 1, as the largest color value is the single node's color.
Graph with a cycle (directed)	Detect cycles using topological sort or DFS and return -1 if a cycle exists, as color values can infinitely increase.
Graph with no edges (isolated nodes)	Return the maximum count of any single color in the colors string.
Graph with all nodes having the same color	The largest color value will be the length of the longest path, handled by considering all paths.
Graph with a long chain (linear graph) of nodes, different colors	The solution should efficiently traverse the chain to find the largest color value.
Large graph (n close to limit) to test scaling	Use topological sort with memoization to avoid recomputation, making it efficient.
Colors string contains characters other than lowercase English letters	Throw an error or return -1, as the input is invalid (problem states lowercase letters).