Count Paths That Can Form a Palindrome in a Tree

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TreesGraphsBit Manipulation

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 the edge between i and parent[i]. s[0] can be ignored.

Return the number of pairs of nodes (u, v) such that u < v and the characters assigned to edges on the path from u to v can be rearranged to form a palindrome.

A string is a palindrome when it reads the same backwards as forwards.

Example 1:

Input: parent = [-1,0,0,1,1,2], s = "acaabc"
Output: 8
Explanation: The valid pairs are:
- All the pairs (0,1), (0,2), (1,3), (1,4) and (2,5) result in one character which is always a palindrome.
- The pair (2,3) result in the string "aca" which is a palindrome.
- The pair (1,5) result in the string "cac" which is a palindrome.
- The pair (3,5) result in the string "acac" which can be rearranged into the palindrome "acca".

Example 2:

Input: parent = [-1,0,0,0,0], s = "aaaaa"
Output: 10
Explanation: Any pair of nodes (u,v) where u < v is valid.

Constraints:

n == parent.length == s.length
1 <= n <= 10⁵
0 <= parent[i] <= n - 1 for all i >= 1
parent[0] == -1
parent represents a valid tree.
s consists of only lowercase 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 maximum number of nodes `n` in the tree? Are there any memory constraints I should be aware of?
If an edge connects a node to itself (a self-loop), is that a valid path to consider, and what character is associated with it?
Is the tree guaranteed to be connected, or could it consist of multiple disjoint components?
Are the edge list and node count always valid, or should I expect and handle potential errors such as an invalid edge list that doesn't correspond to the given number of nodes?
If no palindrome paths exist, what value should I return?

Brute Force Solution

Approach

The brute force method for this problem involves checking every possible path within the tree. For each path, we build the string formed by the node labels and then determine if that string could be rearranged to form a palindrome. We continue exhaustively until every path has been explored and counted if applicable.

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

Start at any node in the tree.
Consider a path that just consists of that single starting node.
Check if the string formed by just that node's label can form a palindrome (a single character always can). If so, increase our count.
Now, extend that path to one of its neighbors.
Build the string formed by concatenating the labels of the nodes in this new path.
Check if the string can form a palindrome. If so, increase the count.
Keep extending the path by adding a neighbor at the end, creating a new string, and checking for palindromes. Repeat this until you reach a dead end (a leaf node).
Once you've exhausted all paths starting from a given node, repeat the entire process by starting from a different node in the tree.
Continue doing this for every single node in the tree as a starting point.
In the end, sum up all the instances where a path formed a string that can become a palindrome. That sum is the final answer.

Code Implementation

def count_palindrome_paths_brute_force(tree):

    number_of_nodes = len(tree)
    palindrome_path_count = 0

    # Iterate through all possible pairs of nodes to check paths
    for start_node in range(number_of_nodes):
        for end_node in range(start_node, number_of_nodes):

            path = find_path(tree, start_node, end_node)

            if path:
                character_counts = {}
                for node_value in path:
                    if node_value not in character_counts:
                        character_counts[node_value] = 0
                    character_counts[node_value] += 1

                odd_count = 0
                for count in character_counts.values():
                    if count % 2 != 0:
                        odd_count += 1

                # Check if the path can form a palindrome
                if odd_count <= 1:
                    palindrome_path_count += 1

    return palindrome_path_count

def find_path(tree, start_node, end_node):
    if start_node == end_node:
        return [start_node]

    paths = {
        start_node: [start_node]
    }

    queue = [start_node]

    while queue:
        current_node = queue.pop(0)

        for neighbor in tree[current_node]:
            if neighbor not in paths:
                paths[neighbor] = paths[current_node] + [neighbor]
                if neighbor == end_node:
                    return paths[neighbor]
                queue.append(neighbor)

    return None

# Example Usage (assuming tree is an adjacency list)
# tree = {
#     0: [1, 2],
#     1: [0, 3, 4],
#     2: [0, 5],
#     3: [1],
#     4: [1],
#     5: [2]
# }
# result = count_palindrome_paths_brute_force(tree)
# print(result)

Big(O) Analysis

Time Complexity

O(n^2 * k) – The brute force approach considers every possible path in the tree. In a tree with n nodes, there can be approximately n^2 paths, as each node can potentially be connected to every other node. For each of these n^2 paths, we need to construct the string formed by concatenating the node labels along the path. Assuming the average path length is proportional to n, and each node label has length k, constructing the string takes O(n * k) time. Then for each path (once the string is built) we need to check if the string can form a palindrome. Forming the path takes at most O(n*k) and palindrome check can be done in O(n*k) time. Finally, after string creation, we can perform the palindromic check in O(n) or O(n*k) time (checking character counts). So, the palindrome check becomes absorbed by the string construction. Therefore, the overall time complexity is O(n^2 * k) where n is the number of nodes and k is the length of the string at each node.

Space Complexity

O(N) – The brute force approach explores paths in the tree using recursion. The maximum depth of the recursion can be N in the worst-case scenario where the tree is a skewed tree (similar to a linked list), where N is the number of nodes in the tree. Each recursive call adds a frame to the call stack. The space needed for the call stack then becomes proportional to the depth of recursion. Thus, the auxiliary space complexity is O(N).

Optimal Solution

Approach

To efficiently count palindrome-forming paths in a tree, we leverage a clever trick using bit manipulation. The core idea is to represent the path's character composition as a bitmask and use this information to quickly check for palindromes.

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

Think of each path from the root to a node as a string of characters, and we want to know how many of these paths can rearrange to form a palindrome.
Instead of actually rearranging characters, we use a 'bitmask'. Each letter of the alphabet will have its own assigned bit in the mask. As we travel down a path, whenever we see a letter, we flip its corresponding bit in the mask.
A palindrome can have at most one letter that appears an odd number of times. This means in our bitmask representation, at most one bit can be set to 1.
So, for each node in the tree, we store the bitmask representing the path from the root to that node.
Now, we use a technique such as depth-first search to traverse the tree. When we visit a node, we check how many paths from the root to other nodes have a bitmask that differs from the current node's bitmask by at most one bit being set.
Each path from the root to another node that has a different bitmask by at most one set bit can be rearranged into a Palindrome. Add that to our count.
By using the bitmask, we have avoided the need to actually construct and check for Palindromes, which makes the process much faster. This enables efficient counting of palindrome-forming paths throughout the entire tree.

Code Implementation

def count_palindrome_paths(number_of_nodes, edges, string):

    graph = [[] for _ in range(number_of_nodes)]
    for node_1, node_2 in edges:
        graph[node_1].append(node_2)
        graph[node_2].append(node_1)

    count = 0

    def depth_first_search(current_node, parent_node, path_bitmask):
        nonlocal count
        path_bitmask ^= (1 << (ord(string[current_node]) - ord('a')))

        # Count paths ending at the current node
        if is_palindrome(path_bitmask):
            count += 1

        # Explore children
        for neighbor_node in graph[current_node]:
            if neighbor_node != parent_node:

                # Traverse only if node is not parent
                depth_first_search(neighbor_node, current_node, path_bitmask)

    def is_palindrome(bitmask):

        # Allows one odd frequency
        if bitmask == 0 or (bitmask & (bitmask - 1)) == 0:
            return True

        return False

    # Initialize bitmask for root node path
    for start_node in range(number_of_nodes):
        depth_first_search(start_node, -1, 0)

    return count

Big(O) Analysis

Time Complexity

O(n) – The solution uses Depth-First Search (DFS) to traverse the tree. In a tree with n nodes, DFS visits each node and edge exactly once. For each node, a constant number of bitwise operations are performed to update and check the bitmask. The cost of DFS traversal is O(n), and since the bitmask operations take constant time, the overall time complexity is dominated by the tree traversal, resulting in O(n).

Space Complexity

O(N) – The depth-first search traversal of the tree utilizes a recursion stack. In the worst-case scenario, the tree could be a skewed tree, resulting in a recursion depth proportional to the number of nodes, N. Additionally, each node stores a bitmask representing the path from the root, but since the bitmask has a fixed size (depending on the alphabet size which is constant), it contributes a constant amount of space per node. Therefore, the primary space complexity comes from the recursion stack, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty tree (n=0 or edges list is empty)	Return 0 as there are no paths.
Tree with only one node (n=1)	Return 1 as the path from the root to itself is considered a palindrome.
Tree where all edge characters are the same	Many paths will be palindromes, test efficiency of palindrome check.
Tree where no paths are palindromes	The algorithm should correctly return 0 without errors.
Large tree (large n) and deep paths	Ensure algorithm avoids excessive recursion depth (stack overflow) and is time-efficient (e.g., using efficient path tracking and palindrome checking).
Integer overflow when calculating path representation (using bitmask)	Consider using a larger integer type or an alternative representation if the number of distinct characters is large enough to cause an overflow.
Disconnected graph (not a single tree)	The problem assumes a tree so handle by ensuring the code does not error out or return incorrect results.
Edge list contains invalid characters (not lowercase letters)	Handle by either validating the inputs before processing or ensuring that any unexpected characters are ignored during the path evaluation.

Empty tree (n=0 or edges list is empty)

How to Handle:

Return 0 as there are no paths.

Tree with only one node (n=1)

How to Handle:

Return 1 as the path from the root to itself is considered a palindrome.

Tree where all edge characters are the same

How to Handle:

Many paths will be palindromes, test efficiency of palindrome check.

Tree where no paths are palindromes

How to Handle:

The algorithm should correctly return 0 without errors.

Large tree (large n) and deep paths

How to Handle:

Ensure algorithm avoids excessive recursion depth (stack overflow) and is time-efficient (e.g., using efficient path tracking and palindrome checking).

Integer overflow when calculating path representation (using bitmask)

How to Handle:

Consider using a larger integer type or an alternative representation if the number of distinct characters is large enough to cause an overflow.

Disconnected graph (not a single tree)

How to Handle:

The problem assumes a tree so handle by ensuring the code does not error out or return incorrect results.

Edge list contains invalid characters (not lowercase letters)

How to Handle:

Handle by either validating the inputs before processing or ensuring that any unexpected characters are ignored during the path evaluation.