Count Paths That Can Form a Palindrome in a Tree

Count Paths That Can Form a Palindrome in a Tree

What is this problem about?

The "Count Paths That Can Form a Palindrome in a Tree interview question" is a high-level algorithmic challenge combining graph theory and bitwise logic. You are given a tree where each edge has a character label ('a'-'z'). Your task is to count the number of paths $(u, v)$ such that the characters on the path can be rearranged to form a palindrome. A set of characters can form a palindrome if and only if at most one character appears an odd number of times.

Why is this asked in interviews?

Companies like Uber and TikTok use the "Count Paths That Can Form a Palindrome in a Tree coding problem" to test a candidate's ability to simplify path-based constraints. It requires the insight that the path between $u$ and $v$ can be represented using their paths to the root. It evaluates knowledge of "Bit Manipulation interview pattern" (specifically bitmasks for parity) and "Depth-First Search" optimization.

Algorithmic pattern used

This problem follows the Bitmask Parity and Path-to-Root pattern.

1. Mask Representation: Represent the parity of character counts using a 26-bit integer (bitmask). If the character 'a' appears an odd number of times, the $0^{th}$ bit is 1.
2. Path Property: The mask of the path between $u$ and $v$ is mask(u) ^ mask(v). This works because XORing with the root path removes the common ancestor path (since $x \oplus x = 0$).
3. Palindrome Condition: A path mask can form a palindrome if mask(u) ^ mask(v) is either 0 or has exactly one bit set.
4. Implementation:
   • Perform a DFS to calculate the mask(root, node) for every node.
   • Use a hash map to store the frequencies of these masks.
   • For each mask $M$ in the map, count pairs within $M$ ($M \oplus M = 0$) and pairs between $M$ and $M \oplus (1 \ll i)$ for $i \in [0, 25]$.

Example explanation

Edge labels: (0-1: 'a', 0-2: 'b')

• mask(0) = 0
• mask(1) = 1 (binary ...001)
• mask(2) = 2 (binary ...010) Check paths:
• (0, 1): mask(0)^mask(1) = 1. Palindrome OK.
• (0, 2): mask(0)^mask(2) = 2. Palindrome OK.
• (1, 2): mask(1)^mask(2) = 3 (binary ...011). Two characters ('a' and 'b') appear once. Not a palindrome. Result: 2 valid paths.

Common mistakes candidates make

• Simulating Paths: Trying to calculate the character counts for every pair $(u, v)$ individually, which is $O(N^2)$. The bitmask approach is $O(N imes 26)$.
• XOR Logic: Failing to realize that the path parity between any two nodes is just the XOR of their root-to-node parities.
• Deduplication: Not handling the $(u, v)$ vs $(v, u)$ pairs correctly in the final count.

Interview preparation tip

Whenever a problem mentions "rearranging to form a palindrome," immediately think of parity bitmasks. For trees, remember the "root-path" XOR trick to find path properties in $O(1)$ after a single DFS traversal.