Longest Path With Different Adjacent Characters

Hard

Amazon

4 views

Topics:

TreesGraphsRecursion

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.

Example 1:

Consider a tree where parent = [-1,0,0,1,1,2] and s = "abacbe".

   0 (a)
  / \
 1 (b) 2 (a)
/ \   \
3 (c) 4 (b) 5 (e)

The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.

Example 2:

Consider a tree where parent = [-1,0,0,0] and s = "aabc"

   0 (a)
 / | \
1 (a) 2 (b) 3 (c)

The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

Write a function to efficiently find the longest path in the tree that meets the specified conditions.

Longest Path With Different Adjacent Characters

Hard

Amazon

4 views

Topics:

TreesGraphsRecursion

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.

Example 1:

Consider a tree where parent = [-1,0,0,1,1,2] and s = "abacbe".

   0 (a)
  / \
 1 (b) 2 (a)
/ \   \
3 (c) 4 (b) 5 (e)

The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.

Example 2:

Consider a tree where parent = [-1,0,0,0] and s = "aabc"

   0 (a)
 / | \
1 (a) 2 (b) 3 (c)

The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

Write a function to efficiently find the longest path in the tree that meets the specified conditions.

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 are the constraints on `n` (the number of nodes)?
Can the `parent` array contain any invalid values, such as a parent index that's out of bounds or a cycle involving node 0 (the root)?
Is the string `s` guaranteed to only contain lowercase English letters, or can it contain other characters?
If the tree is empty (n=0), what should I return?
If there are multiple longest paths satisfying the condition, should I return the length of any one of them?

Brute Force Solution

Approach

To find the longest path, we'll explore every single possible path in the given structure. We'll start from each point, trace every route, and see which route is the longest while obeying the rule about different characters.

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

Pick any starting point.
From that starting point, consider every possible direction you can go.
For each of those directions, check if the next point has a different character than the current point.
If it does, move to that next point and add it to your path.
Continue this process, exploring all possible paths from your current position, always making sure the next character is different.
When you reach a point where you can't go any further because all adjacent points either have the same character or have already been visited in the current path, stop exploring that path.
Repeat this whole process starting from every single possible starting point.
Keep track of the length of each valid path you find.
After exploring all possible paths from all starting points, compare the lengths of all the paths you found.
The longest path among all of them is your answer.

Code Implementation

def longest_path_with_different_adjacent_characters_brute_force(parent, string):
    number_of_nodes = len(string)
    adjacency_list = [[] for _ in range(number_of_nodes)]
    for index, parent_node in enumerate(parent):
        if index == 0:
            continue
        adjacency_list[parent_node].append(index)
        adjacency_list[index].append(parent_node)

    def depth_first_search(node, visited):
        max_path_length = 0

        # Explore all neighbors
        for neighbor in adjacency_list[node]:
            if neighbor not in visited and string[node] != string[neighbor]:

                # Check if the neighbor has a different character
                new_visited = set(visited)
                new_visited.add(neighbor)
                path_length = depth_first_search(neighbor, new_visited)
                max_path_length = max(max_path_length, path_length)

        return max_path_length + 1

    longest_path = 0
    for start_node in range(number_of_nodes):

        # Try every node as the start
        visited_nodes = {start_node}
        path_length = depth_first_search(start_node, visited_nodes)
        longest_path = max(longest_path, path_length)

    return longest_path

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores every possible path in the graph. In the worst-case scenario, the graph could be structured in such a way that each node is connected to every other node. This means from each starting node, the algorithm might have to visit almost all other nodes to find the longest path. Because it explores every possible path branching out from each node, the number of paths grows factorially with the number of nodes, which leads to O(n!). The algorithm essentially performs a brute-force search of all possible paths.

Space Complexity

O(N) – The algorithm implicitly uses a recursion stack due to the exploration of all possible paths from each starting point. In the worst-case scenario, the depth of the recursion can be proportional to the number of nodes in the graph, represented by N, as the algorithm explores each node. This recursion depth dictates the amount of space used to store the call stack. Therefore, the auxiliary space complexity is O(N), where N is the number of nodes.

Optimal Solution

Approach

The goal is to find the longest path in a tree where no two directly connected locations have the same letter. We can solve this efficiently by exploring the tree in a specific order and keeping track of the best paths we've seen so far at each location.

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

Start at the very bottom of the tree (the leaves).
For each location, figure out the longest path that starts at that location and goes downwards, following the rules about different letters.
When exploring a location, look at all its immediate neighbors further down the tree.
If a neighbor has a different letter, consider including it in the current location's path and update the length.
Keep track of the two best paths found at each location, because the longest path might go 'up' to this location, then 'down' through two different branches.
As you work your way up the tree, combine paths found at the bottom locations to build longer and longer paths, always ensuring neighboring letters are different.
The longest path found at any location in the tree is the final answer.

Code Implementation

def longest_path_with_different_adjacent_characters(parent, text):
    number_of_nodes = len(parent)
    children = [[] for _ in range(number_of_nodes)]
    for index in range(1, number_of_nodes):
        children[parent[index]].append(index)

    longest_path = 1

    def depth_first_search(node_index):
        nonlocal longest_path
        longest_chain_one = 0
        longest_chain_two = 0

        for child_index in children[node_index]:
            chain_length = depth_first_search(child_index)

            # Only consider children with different characters.
            if text[node_index] != text[child_index]:

                if chain_length > longest_chain_one:
                    longest_chain_two = longest_chain_one
                    longest_chain_one = chain_length
                elif chain_length > longest_chain_two:
                    longest_chain_two = chain_length

        # Update the longest path found so far
        longest_path = max(longest_path, longest_chain_one + longest_chain_two + 1)

        # The longest path that can be extended to the parent.
        return longest_chain_one + 1

    depth_first_search(0)
    return longest_path

Big(O) Analysis

Time Complexity

O(n) – The algorithm traverses the tree using a depth-first search (DFS) approach, visiting each node once. For each node, it iterates through its children (neighbors). In a tree, the number of edges is n-1, where n is the number of nodes. Therefore, the total number of neighbor checks across all nodes is proportional to the number of edges, which is O(n). Consequently, the overall time complexity is dominated by the tree traversal, making it O(n).

Space Complexity

O(N) – The algorithm implicitly uses recursion, and in the worst-case scenario, the recursion depth can be proportional to the number of nodes in the tree, denoted as N. The recursion stack stores information about each active function call. Furthermore, the algorithm mentions storing the two best paths found at each location. While not explicitly stated as a dedicated data structure like a hashmap, this 'keeping track' still suggests each node might need some small constant amount of space to remember best path lengths, accumulating to space proportional to the number of nodes. Therefore, the auxiliary space is O(N).

Edge Cases

Case	How to Handle
Empty parent array (n=0)	Return 0, as there are no nodes and therefore no path.
Single node tree (n=1)	Return 1, as a single node constitutes a path of length 1.
All nodes have the same character in string s	The longest path will be a single node, so return 1.
Parent array creates a cycle (not a valid tree)	The algorithm should still function correctly, returning the longest valid path found without being affected by cycles since we are building the tree using parent array and characters.
Large n (number of nodes), potential stack overflow with recursion	Consider using iterative DFS or BFS with an explicit stack to avoid stack overflow.
Null or invalid characters in string s	The algorithm should handle null characters correctly by comparing them like any other characters, and we may need to add a check for other invalid characters as needed to filter errors.
Parent array is malformed (e.g., parent[0] != -1 or node refers to itself)	Handle gracefully and perhaps throw an exception, or the algorithm might create a corrupted tree and return a nonsensical answer.
Maximum tree depth close to n (highly skewed tree)	Algorithm should handle this correctly by traversing each node and finding the longest path.