The Most Similar Path in a Graph

Hard

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GraphsDynamic Programming

We have n cities labeled from 0 to n - 1. Given the array roads where roads[i] = [a_i, b_i] represents a bidirectional road between cities a_i and b_i, and given the array names of size n where names[i] is the name of the city i.

You are planning a trip that starts in city 0 and ends in city n - 1. You want to find a path of length m such that the sequence of cities visited along this path is most similar to a given targetPath.

A path of length m is similar to targetPath if they have the same length and the number of cities that have the same name at each position is maximized. In other words, you want to maximize the number of indices i such that path[i] == targetPath[i] (0 <= i < m).

Return a list of the cities visited on the most similar path. You may return any path if multiple paths have the same similarity.

The graph is guaranteed to be connected, and each node has at most 4 neighbors.

Example 1:

Input: n = 5, roads = [[0,2],[0,3],[1,2],[1,3],[1,4],[2,4]], names = ["ATL","PEK","LAX","DXB","HND"], targetPath = ["ATL","DXB","HND","LAX"]
Output: [0,3,4,2]
Explanation: [0,3,4,2], [0,2,4,2], [0,3,1,4] are the most similar paths.

Example 2:

Input: n = 4, roads = [[1,0],[2,0],[3,0],[2,1],[3,1],[3,2]], names = ["ATL","PEK","LAX","DXB"], targetPath = ["ATL","PEK","LAX","DXB"]
Output: [0,1,2,3]
Explanation: [0,1,2,3] are the only possible paths.

Example 3:

Input: n = 6, roads = [[0,1],[1,2],[2,3],[3,4],[4,5]], names = ["ATL","PEK","LAX","DXB","HND","ORD"], targetPath = ["ATL","DXB","HND","DXB","ORD"]
Output: [0,3,4,3,5]

Constraints:

2 <= n <= 100
1 <= roads.length <= n * (n - 1) / 2
0 <= a_i, b_i <= n - 1
a_i != b_i
The graph is guaranteed to be connected and each node has at most 4 neighbors.
names.length == n
1 <= names[i].length <= 10
names[i] consists of only upper and lower case English letters.
There can be two cities with the same name.
1 <= targetPath.length <= 100
1 <= targetPath[i].length <= 10
targetPath[i] consists of only upper and lower case English letters.

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 format of the graph input? Is it an adjacency list, adjacency matrix, or something else, and what are the possible node labels?
What are the constraints on the length of the `targetPath` and the number of nodes and edges in the graph?
If multiple paths have the same minimum edit distance to the `targetPath`, which one should I return? Is there a preference, such as the lexicographically smallest path?
Can the `targetPath` contain node labels that are not present in the graph, and if so, how should that be handled?
If no path exists in the graph, what should I return? Should I return null, an empty list, or throw an exception?

Brute Force Solution

Approach

The brute force approach to finding the most similar path in a graph involves exploring every possible path in the graph. We generate all potential paths, compare each one to the target path, and select the path that is most similar. It's like trying every single road in a city to find the one closest to your desired route.

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

Start with the first node as the beginning of a possible path.
From that first node, explore all its neighbors, adding each neighbor to the path and creating a new possible path.
Continue this process for each node in the growing paths, exploring all of its neighbors in the graph to expand the paths further.
Keep going until all paths have the same length as the target path or we reach a dead end.
For each complete path, compare it to the target path to calculate a 'difference score' based on how many nodes are different.
After exploring all possible paths, choose the path with the lowest difference score; that is, the path most similar to the target path.

Code Implementation

def find_most_similar_path_brute_force(graph, target_path):
    most_similar_path = None
    minimum_difference = float('inf')
    target_path_length = len(target_path)

    def calculate_difference(path):
        difference = 0
        for index in range(target_path_length):
            if index < len(path) and path[index] != target_path[index]:
                difference += 1
            elif index >= len(path):
                difference += 1 
        return difference

    def explore_paths(current_node, current_path):
        nonlocal most_similar_path, minimum_difference

        current_path.append(current_node)

        if len(current_path) == target_path_length:
            # A complete path is found
            difference = calculate_difference(current_path)

            if difference < minimum_difference:
                minimum_difference = difference
                most_similar_path = current_path.copy()

        elif len(current_path) < target_path_length:
            # Recursively extend the path
            if current_node in graph:
                for neighbor in graph[current_node]:
                    explore_paths(neighbor, current_path.copy())

    # Iterate to begin exploration from each node
    for start_node in graph:
        explore_paths(start_node, [])

    # Ensure a path is returned if graph isn't empty
    if most_similar_path is None and len(graph) > 0:
        first_node = next(iter(graph))
        explore_paths(first_node, [])

    #Handle edge case of an empty graph
    if most_similar_path is None:
      return []

    return most_similar_path

Big(O) Analysis

Time Complexity

O(V * A^L) – The algorithm explores all possible paths in the graph, where V is the number of starting vertices in the graph. From each starting vertex, the algorithm explores paths of length L, where L is the length of the target path. In the worst case, each vertex has A outgoing edges, or an average degree of A. Thus, the number of possible paths that need to be compared against the target path is proportional to A^L. Comparing each possible path against the target path of length L takes O(L) time, but that is dominated by path exploration. Therefore, the overall time complexity is O(V * A^L).

Space Complexity

O(V^L) – The brute force approach explores all possible paths of length L (the target path length) in a graph with V vertices. We are essentially storing each possible path as it is constructed. At each step, we create new paths from the existing ones, leading to exponential growth. Therefore, the auxiliary space is dominated by the memory required to store all these paths, which can be up to V^L in the worst case where every vertex is connected to every other vertex.

Optimal Solution

Approach

The goal is to find a path in a graph that's most similar to a given target path. We'll explore the graph while remembering how closely our current path matches the target, avoiding recalculating similarity for the same locations.

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

Start at each possible starting node in the graph.
Explore the graph one step at a time, comparing each node's name to the corresponding word in the target path.
Remember the 'score' (how similar the current path is to the target path) at each node we visit; if we visit the same node with the same path length, keep the better score.
As we explore, if we reach the end of the target path, record the path we took and its overall similarity score.
After exploring all possible paths from all starting nodes, pick the path with the highest similarity score.

Code Implementation

def most_similar_path(number_of_nodes, graph, target_path):
    best_score = -1
    best_path = []

    # Iterate through all possible starting nodes.
    for start_node in range(number_of_nodes):
        path = [start_node]
        score = 0 if graph[start_node] == target_path[0] else 1
        result_path, result_score = dfs(
            graph,
            target_path,
            start_node,
            1,
            path,
            score,
            number_of_nodes
        )

        if result_score > best_score:
            best_score = result_score
            best_path = result_path

    return best_path

def dfs(graph_nodes, target, current_node, target_index, current_path, current_score, number_of_nodes):
    if target_index == len(target):
        return current_path, current_score

    best_path = current_path
    best_score = -1

    # Explore neighbors of the current node.
    for neighbor in graph_nodes[current_node]:
        new_score = current_score + (0 if graph_nodes[neighbor] == target[target_index] else 1)
        new_path = current_path + [neighbor]

        # Recursive call to explore further down the path
        path, score = dfs(
            graph_nodes,
            target,
            neighbor,
            target_index + 1,
            new_path,
            new_score,
            number_of_nodes
        )

        # Update best path if a better one is found.
        if score > best_score:
            best_score = score
            best_path = path

    return best_path, best_score

def build_graph(paths, node_names):
    graph = {}
    for i, node_name in enumerate(node_names):
        graph[i] = node_name
    graph_nodes = [[] for _ in range(len(node_names))]
    for path in paths:
        graph_nodes[path[0]].append(path[1])
        graph_nodes[path[1]].append(path[0])
    return graph_nodes

Big(O) Analysis

Time Complexity

O(N * M * 2^M) – Let N be the number of nodes in the graph and M be the length of the target path. The algorithm explores all possible paths starting from each node. In the worst case, we might need to visit each node for each possible length of the target path (up to length M). At each node, we explore its neighbors. Since graph exploration can potentially create exponential branching, particularly if the graph is dense, the number of paths could grow exponentially with M (the depth of the search), approximated by 2^M. Therefore, we visit N nodes * each for up to M path lengths * with a branching factor that can grow to 2^M which is multiplied by the number of edges M to reach each neighbour. Thus the time complexity is O(N * M * 2^M).

Space Complexity

O(M * N) – The dominant space usage comes from remembering the 'score' at each node visited for a given path length. This requires storing the best score seen so far for each node up to the length of the target path. Let M be the length of the target path and N be the number of nodes in the graph. We are essentially storing a matrix of size M x N to memoize the best similarity scores for each node and path length, resulting in O(M * N) space. The recording of the most similar path would take space proportional to the length of the target path, M, which is less than M * N. Therefore the total auxiliary space complexity is O(M * N).

Edge Cases

Case	How to Handle
Empty graph (no nodes or edges)	Return an empty path since there's no path to find.
Null or empty target path	Return an empty path or the best possible path from a starting node depending on problem specifications
Graph with only one node	Return the single node as the path if the target path's first node matches, otherwise return an empty path.
Target path longer than the longest possible path in the graph	Return the most similar path found up to the maximum path length in the graph.
Cyclic graph with a very long cycle	Implement cycle detection in the search to prevent infinite loops, potentially using a visited set.
Multiple equally similar paths exist	Return any one of the most similar paths or specify a criteria for selection (e.g., lexicographically smallest).
Node names/labels contain special characters or are extremely long strings	Ensure node name comparisons are handled correctly and limit node name length for memory safety.
Graph is disconnected and the target path exists only in one component	Iterate through possible starting nodes in all components to find the component containing the best matching path.

Empty graph (no nodes or edges)

How to Handle:

Return an empty path since there's no path to find.

Null or empty target path

How to Handle:

Return an empty path or the best possible path from a starting node depending on problem specifications

Graph with only one node

How to Handle:

Return the single node as the path if the target path's first node matches, otherwise return an empty path.

Target path longer than the longest possible path in the graph

How to Handle:

Return the most similar path found up to the maximum path length in the graph.

Cyclic graph with a very long cycle

How to Handle:

Implement cycle detection in the search to prevent infinite loops, potentially using a visited set.

Multiple equally similar paths exist

How to Handle:

Return any one of the most similar paths or specify a criteria for selection (e.g., lexicographically smallest).

Node names/labels contain special characters or are extremely long strings

How to Handle:

Ensure node name comparisons are handled correctly and limit node name length for memory safety.

Graph is disconnected and the target path exists only in one component

How to Handle:

Iterate through possible starting nodes in all components to find the component containing the best matching path.