Palindromic Path Queries in a Tree - Solution & Explanation

Q: Is Palindromic Path Queries in a Tree easy or hard?

This problem is typically rated Hard because it combines multiple advanced concepts: tree traversal, LCA preprocessing, bitmask parity checks, and efficient path queries. Understanding how XOR cancels shared prefixes along tree paths is the key insight.

Q: Palindromic Path Queries in a Tree Python/Java solution

Most implementations compute prefix parity masks during DFS and store ancestor tables for LCA. Queries combine masks using XOR and check if the result has at most one set bit. The same algorithm translates easily across Python, Java, C++, and Go because it mainly relies on bit operations and adjacency lists.

Q: What is the best approach for Palindromic Path Queries in a Tree?

The most efficient approach uses prefix bitmasks combined with Lowest Common Ancestor (LCA). Each node stores a parity bitmask representing character counts from the root to that node. For a query between two nodes, XOR the masks and adjust using the LCA. The resulting mask reveals whether at most one character has an odd frequency, which determines if the path can form a palindrome.

Q: How to solve Palindromic Path Queries in a Tree in O((n+q) log n)?

Run a DFS from the root to compute prefix parity bitmasks for each node. Precompute LCA using binary lifting. For a query between nodes u and v, compute mask = mask[u] XOR mask[v] XOR bit(char at lca). The path forms a palindrome if mask & (mask - 1) equals zero, meaning at most one bit is set.

Q: Is Palindromic Path Queries in a Tree asked at Google/Amazon/Meta?

Tree path queries combined with bitmask tricks and LCA frequently appear in interviews at companies like Google, Amazon, and Meta. Variants include checking palindrome paths, counting characters on tree paths, and parity-based queries using bit manipulation.

Q: What data structure is used in Palindromic Path Queries in a Tree?

Key data structures include trees for representing the graph, binary lifting tables for LCA computation, bitmasks for tracking character parity, and optionally segment trees or Heavy-Light Decomposition for advanced path queries or updates.

Q: What is the time complexity of Palindromic Path Queries in a Tree?

Using prefix bitmasks with LCA preprocessing runs in O(n log n) for building the tree structure and O(log n) per query. Total complexity becomes O((n + q) log n). The palindrome check itself is constant time because it only requires verifying that the resulting bitmask has at most one set bit.

HardArray String Divide and Conquer Tree5 min read

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Problem Statement

You are given an undirected tree with n nodes labeled 0 to n - 1. This is represented by a 2D array edges of length n - 1, where edges[i] = [u_i, v_i] indicates an undirected edge between nodes u_i and v_i.

You are also given a string s of length n consisting of lowercase English letters, where s[i] represents the character assigned to node i.

You are also given a string array queries, where each queries[i] is either:

"update u_i c": Change the character at node u_i to c. Formally, update s[u_i] = c.
"query u_i v_i": Determine whether the string formed by the characters on the unique path from u_i to v_i (inclusive) can be rearranged into a palindrome.

Return a boolean array answer, where answer[j] is true if the j^th query of type "query u_i v_i" can be rearranged into a palindrome, and false otherwise.

Example 1:

Input: n = 3, edges = [[0,1],[1,2]], s = "aac", queries = ["query 0 2","update 1 b","query 0 2"]

Output: [true,false]

Explanation:

"query 0 2": Path 0 → 1 → 2 gives "aac", which can be rearranged to form "aca", a palindrome. Thus, answer[0] = true.
"update 1 b": Update node 1 to 'b', now s = "abc".
"query 0 2": Path characters are "abc", which cannot be rearranged to form a palindrome. Thus, answer[1] = false.

Thus, answer = [true, false].

Example 2:

Input: n = 4, edges = [[0,1],[0,2],[0,3]], s = "abca", queries = ["query 1 2","update 0 b","query 2 3","update 3 a","query 1 3"]

Output: [false,false,true]

Explanation:

"query 1 2": Path 1 → 0 → 2 gives "bac", which cannot be rearranged to form a palindrome. Thus, answer[0] = false.
"update 0 b": Update node 0 to 'b', now s = "bbca".
"query 2 3": Path 2 → 0 → 3 gives "cba", which cannot be rearranged to form a palindrome. Thus, answer[1] = false.
"update 3 a": Update node 3 to 'a', s = "bbca".
"query 1 3": Path 1 → 0 → 3 gives "bba", which can be rearranged to form "bab", a palindrome. Thus, answer[2] = true.

Thus, answer = [false, false, true].

Constraints:

1 <= n == s.length <= 5 * 10⁴
edges.length == n - 1
edges[i] = [u_i, v_i]
0 <= u_i, v_i <= n - 1
s consists of lowercase English letters.
The input is generated such that edges represents a valid tree.
1 <= queries.length <= 5 * 10⁴
- queries[i] = "update u_i c" or
- queries[i] = "query u_i v_i"
- 0 <= u_i, v_i <= n - 1
- c is a lowercase English letter.

Approach Overview

Problem Overview: You are given a tree where each node contains a character. For multiple queries, determine whether the characters along the path between two nodes can be rearranged to form a palindrome. A path is palindromic if at most one character appears an odd number of times.

Approach 1: Brute Force Path Collection (O(n) per query, O(n) space)

For every query, explicitly find the path between the two nodes using DFS or by computing the path through their Lowest Common Ancestor (LCA). Collect all characters along that path, count their frequencies using a hash map, and check if at most one character has an odd count. This approach is straightforward but inefficient because each query may traverse up to O(n) nodes. With many queries, the total runtime becomes O(n * q).

Approach 2: Prefix Bitmask with LCA (O((n + q) log n) time, O(n log n) space)

Encode character frequency parity using a bitmask. Each node stores a mask where the i-th bit represents whether the count of character i from the root to that node is odd. Compute these masks during a DFS. For a query between nodes u and v, compute the mask for the path using mask[u] XOR mask[v] XOR char(lca). A path can form a palindrome if the resulting mask has at most one bit set. LCA is computed using binary lifting on the tree. Checking the palindrome condition becomes a constant-time bit operation.

Approach 3: Heavy-Light Decomposition with Segment Tree (O((n + q) log n) time, O(n) space)

When queries require flexible path aggregation or potential updates, decompose the tree using Heavy-Light Decomposition. Each chain maps to an array segment, and a segment tree stores bitmasks representing character parity. For each query, split the path into O(log n) segments and combine masks using XOR. The final mask is checked for the single-bit condition. This approach combines tree decomposition with segment tree range queries and fits naturally with divide and conquer style tree processing.

Recommended for interviews: The prefix bitmask with LCA approach is usually expected. It demonstrates understanding of tree traversal, bit manipulation, and efficient path queries. Brute force confirms you understand the palindrome condition, but the optimized solution shows you can reduce frequency tracking to a constant-time parity check.

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Detailed Complexity Analysis

Approach	Time	Space	When to Use
Brute Force Path Traversal	O(n) per query	O(n)	Useful for understanding the problem or when the number of queries is very small
Prefix Bitmask + LCA	O((n + q) log n)	O(n log n)	Best general solution for static trees with many queries
Heavy-Light Decomposition + Segment Tree	O((n + q) log n)	O(n)	Preferred when path queries combine with updates or advanced tree queries

Video Solution

Q4. Palindromic Path Queries in a Tree || LCA, Tree, Binary Lifting || Leetcode Biweekly 176 || 2X 🚀 • Rajan Keshari ( CSE - IIT Dhanbad ) • 863 views views

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Frequently Asked Questions

Is Palindromic Path Queries in a Tree easy or hard?

This problem is typically rated Hard because it combines multiple advanced concepts: tree traversal, LCA preprocessing, bitmask parity checks, and efficient path queries. Understanding how XOR cancels shared prefixes along tree paths is the key insight.

Palindromic Path Queries in a Tree Python/Java solution

Most implementations compute prefix parity masks during DFS and store ancestor tables for LCA. Queries combine masks using XOR and check if the result has at most one set bit. The same algorithm translates easily across Python, Java, C++, and Go because it mainly relies on bit operations and adjacency lists.

What is the best approach for Palindromic Path Queries in a Tree?

The most efficient approach uses prefix bitmasks combined with Lowest Common Ancestor (LCA). Each node stores a parity bitmask representing character counts from the root to that node. For a query between two nodes, XOR the masks and adjust using the LCA. The resulting mask reveals whether at most one character has an odd frequency, which determines if the path can form a palindrome.

How to solve Palindromic Path Queries in a Tree in O((n+q) log n)?

Run a DFS from the root to compute prefix parity bitmasks for each node. Precompute LCA using binary lifting. For a query between nodes u and v, compute mask = mask[u] XOR mask[v] XOR bit(char at lca). The path forms a palindrome if mask & (mask - 1) equals zero, meaning at most one bit is set.

Is Palindromic Path Queries in a Tree asked at Google/Amazon/Meta?

Tree path queries combined with bitmask tricks and LCA frequently appear in interviews at companies like Google, Amazon, and Meta. Variants include checking palindrome paths, counting characters on tree paths, and parity-based queries using bit manipulation.

What data structure is used in Palindromic Path Queries in a Tree?

Key data structures include trees for representing the graph, binary lifting tables for LCA computation, bitmasks for tracking character parity, and optionally segment trees or Heavy-Light Decomposition for advanced path queries or updates.

What is the time complexity of Palindromic Path Queries in a Tree?

Using prefix bitmasks with LCA preprocessing runs in O(n log n) for building the tree structure and O(log n) per query. Total complexity becomes O((n + q) log n). The palindrome check itself is constant time because it only requires verifying that the resulting bitmask has at most one set bit.

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Problem Info

DifficultyHard

Acceptance37.3%

Approaches0

Reading time5 min