Problem Description

In this problem, we are given a tree with n nodes, where each node is numbered from 0 to n - 1. The tree is rooted at node 0, and we are provided a parent array named parent that defines the tree's structure. The parent[i] value represents the parent node of node i, and for the root node 0, we have parent[0] == -1 because it does not have any parent.

Along with the tree, we have a string s of length n that assigns characters to the edges in the tree. The character s[i] corresponds to the edge between node i and its parent parent[i]. Note that s[0] can be ignored because it is associated with the root, which has no parent.

The task is to find the number of distinct pairs of nodes (u, v) where u < v and the series of characters assigned to the edges in the path from u to v can be rearranged to form a palindrome. A palindrome is a sequence that reads the same backward and forward.

Flowchart Walkthrough

To analyze Leetcode 2791, "Count Paths That Can Form a Palindrome in a Tree," using the provided algorithm flowchart, let's go through the decision nodes in the flowchart to determine the appropriate algorithm:

1. Is it a graph?

   • Yes: The problem involves a tree, which is a specialized type of graph with no cycles and is connected.

2. Is it a tree?

   • Yes: The structure described in the problem is explicitly mentioned as a tree.

3. DFS:

   • Since it is a tree, Depth-First Search (DFS) is typically used for tree-related problems to explore all paths or perform operations on paths, such as checking for palindrome conditions from any node to any other node.

The flowchart confirms that DFS is the suitable choice for solving this problem as the tree structure lends itself well to recursive traversal techniques like DFS, which can efficiently explore all possible paths for checking palindrome conditions.

For a visual representation and further understanding of the decision-making process, please refer to the Flowchart.

Intuition

To solve this problem, we need to find all paths in the tree that can create palindromes when the edge characters are rearranged. However, since directly checking every path would be too slow, we approach the problem using a bitwise XOR technique.

We begin with the observation that, for a string of characters to be rearranged into a palindrome, each character must appear an even number of times except for at most one character which can be in the middle. In terms of edges in the path, it means that every character should connect an even number of edges except for at most one.

Translating this idea into an algorithm, we use a bitwise representation to keep track of even and odd counts of edge characters along any path. We create a 26-bit integer to represent the count of each letter in the alphabet, where each bit corresponds to a specific character. If a bit is set, it means an odd count of that character; if it is not set, it means an even count.

Now, through depth-first search (DFS), we traverse the tree starting at the root node and calculate the XOR value representing the character count at each node. For each node, we update the global answer (ans) by counting how many times we've seen this XOR pattern (since the path from the root to this node has the same palindrome check as some previous path), and how many times we've seen this XOR pattern with exactly one bit flipped (since that represents just a single character being odd, which is allowable for palindromes).

The overall process includes accumulating the XOR value while descending through the tree, checking palindromic conditions at each step, and updating the global count of valid paths.

The provided solution keeps track of the counts using a Counter dictionary (cnt), which maps each 26-bit integer to the number of times that bit pattern has previously occurred during the DFS. The dfs() function is called recursively to traverse the tree, and the ans variable, defined as nonlocal, is used to accumulate the total number of palindromic paths.

Learn more about Tree, Depth-First Search, Dynamic Programming and Bitmask patterns.

Solution Approach

The solution uses a depth-first search (DFS) algorithm to traverse the tree, taking advantage of a bitwise representation to encode the frequency of edge characters along the path from the root to any given node.

Data Structures Used:

• A defaultdict (imported from collections module) named g to represent the adjacency list of the tree, where each node has a list of tuples containing its child node and the corresponding edge character's bit representation.
• A Counter (also from collections module) named cnt to keep track of the number of times we've seen a particular 26-bit representation of character frequencies.

The DFS Algorithm and Bitwise Patterns:

Within the provided code, the dfs() function is responsible for exploring the tree:

• It accepts two arguments: i, the current node index, and xor, the sum of edge character bit representations up to this node.
• ans is a nonlocal variable that accumulates the number of valid paths that can form a palindrome. It is incremented both when an existing xor value is found in cnt and when a single-bit difference version of xor is found in cnt.
• As the function traverses the tree, it calculates the new xor value for each child by using the bitwise XOR operation (^) between the current xor and the bit representation of the child edge's character.
• A loop through 26 different bit flips (for k in range(26):) is used to check if there's a valid path that has exactly one character with an odd count, which is still acceptable for palindrome formation. If such a condition is found, ans is also incremented for these occurrences.
• The cnt dictionary keeps track of how many times we've seen this particular xor pattern by incrementing cnt[x].
• The recursion continues by calling dfs(j, x) for each child j.

Building the Bit Representation:

• For each character in the edge, a 26-bit integer is built by shifting 1 to the left ord(s[i]) - ord('a') times, which indicates the position of the character in the alphabet (e.g., 'a' corresponds to shifting 0 times as it's the first letter, 'b' corresponds to shifting 1 time, and so on).
• This way, each bit of the integer represents the count's parity of a corresponding character on the path from the root to the current node.

Initialization and Invocation:

• The function countPalindromePaths initializes the adjacency list g and the Counter cnt with {0: 1}, which corresponds to the bit representation of an empty path (all bits set to zero, meaning even count for every character).
• It then calls dfs(0, 0), starting the traversal from the root node with an initial xor of 0.

By using these techniques, the code efficiently finds all palindromic paths within the tree.

Example Walkthrough

Let's use a small example to illustrate the solution approach. Consider the following tree with n = 4 nodes and its associated parent array and string:

parent = [-1, 0, 1, 1]
s = "abca"

This defines a tree where:

• Node 0 is the root as parent[0] == -1.
• Node 1 is the child of node 0.
• Nodes 2 and 3 are children of node 1.

The string s colors the edges as follows:

• Edge from node 1 to node 0 (the parent) with character 'a'.
• Edge from node 2 to node 1 with character 'b'.
• Edge from node 3 to node 1 with character 'c'.

Thus, the tree looks like this:

    0
   / 
  a  
 /
1
 / \
b   c
/     \
2       3

In the bitwise representation:

• For edge 'a' (from 0 to 1), the bit is 1 << (ord('a') - ord('a')) = 1 << 0 = 1.
• For edge 'b' (from 1 to 2), the bit is 1 << (ord('b') - ord('a')) = 1 << 1 = 2.
• For edge 'c' (from 1 to 3), the bit is 1 << (ord('c') - ord('a')) = 1 << 2 = 4.

Now, we'll walk through the DFS algorithm with this example:

1. We start at the root node 0 with an initial xor value of 0.

2. We visit the child node 1. Edge 0-1 is labeled 'a', so the xor value becomes 0 ^ 1 = 1.

3. At node 1, there are two children. We explore each of them in turn:

   • Visiting node 2, the edge 1-2 is 'b', so the xor becomes 1 ^ 2 = 3. We now have a path 0 - 1 - 2 with edge characters 'a', 'b', which gives a bitwise pattern of 011 telling us the count of character 'a' is odd, 'b' is odd, and 'c' is even (hence not set).
   • Visiting node 3, the edge 1-3 is 'c', so the xor becomes 1 ^ 4 = 5. The path 0 - 1 - 3 has edge characters 'a', 'c', resulting in a bitwise pattern of 101.

4. For each of these nodes 2 and 3, we check the presence of the xor in cnt and check for each possible single-bit flip.

   • When visiting node 2, the xor value before visiting children is 3, which has not been seen before, so cnt[3] is set to 1.
   • For node 3, similarly, the xor value is 5, and cnt[5] is set to 1.
   • There are no single-bit difference patterns for 3 or 5 in cnt yet, so ans stays the same for now.

5. We do not increment the global ans during this initial pass as the DFS is still in progress.

6. Finally, after traversing the entire tree, we find that we have the following pair that can form a palindrome:

   • Nodes (2, 3): The path characters are 'b' and 'c'. By reordering the characters, we can't form a palindrome as there must be an even count of each character or an even count of all characters except one.

At the end of our DFS, we find that there are no valid palindromic paths in this particular example, so the answer (ans) is 0.

Solution Implementation

Time and Space Complexity

The given Python code defines a method countPalindromePaths that returns the number of palindromic paths in a tree-like structure, where each node is labeled with a character from string s. The tree structure is defined by the parent list, where parent[i] is the parent of the i-th node.

Time Complexity

The time complexity of the code is O(N * 26), where N is the number of nodes in the tree. Here's the breakdown:

1. The dfs function is called recursively for each node exactly once, resulting in N calls.
2. Inside the dfs function, for each node i, it iterates over all its children followed by a loop running 26 times (for each lowercase alphabet letter).
3. Every step inside the inner for loop has constant time operations except for the recursive dfs call that happens once for each child node.

Therefore, we have a single dfs call per node and a fixed number of additional iterations per node, resulting in a time complexity of O(N * 26).

Space Complexity

The space complexity of the code is O(N), where N is the number of nodes in the tree. Here's why:

1. The g dictionary and cnt counter have at most N entries, as they store vertices and the number of times certain XOR values occur.
2. The dfs call stack can grow up to O(N) in the case of a tree that degenerates into a linked list.
3. The other variables like x and ans use constant space.

As a result, the total space complexity is determined by the size of the tree and bookkeeping variables, leading to O(N).

In conclusion, the given Python code has a time complexity of O(N * 26) and a space complexity of O(N).

Learn more about how to find time and space complexity quickly using problem constraints.