Palindromic Path Queries in a Tree solution and explanation
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The problem
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] = [ui, vi] indicates an undirected edge between nodes ui and vi.
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 ui c": Change the character at nodeuitoc. Formally, updates[ui] = c."query ui vi": Determine whether the string formed by the characters on the unique path fromuitovi(inclusive) can be rearranged into a palindrome.
Return a boolean array answer, where answer[j] is true if the jth query of type "query ui vi" 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": Path0 → 1 → 2gives"aac", which can be rearranged to form"aca", a palindrome. Thus,answer[0] = true."update 1 b": Update node 1 to'b', nows = "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": Path1 → 0 → 2gives"bac", which cannot be rearranged to form a palindrome. Thus,answer[0] = false."update 0 b": Update node 0 to'b', nows = "bbca"."query 2 3": Path2 → 0 → 3gives"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": Path1 → 0 → 3gives"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 * 104edges.length == n - 1edges[i] = [ui, vi]0 <= ui, vi <= n - 1sconsists of lowercase English letters.- The input is generated such that
edgesrepresents a valid tree. 1 <= queries.length <= 5 * 104queries[i] = "update ui c"orqueries[i] = "query ui vi"0 <= ui, vi <= n - 1cis 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.
Solution
Thinking
A path can be rearranged into a palindrome iff at most one character has an odd count. With
nand the number of queries at most5 times 10^4, walking each path is too slow.The bottleneck is supporting point updates while still reading the parity of every letter on an arbitrary path.
Pack the
26letter parities into a bitmask. The path mask is the XOR of the two rootward prefixes, then the XOR of the LCA's character: the shared prefix and the LCA itself both cancel. Writing each node's mask at its Euler enter and exit times makes a prefix of an XOR Fenwick tree equal the rootward mask.Binary lifting finds the LCA. The path is palindromic when the mask has at most one bit set. An update XORs the difference of the old and new masks at that node's enter and exit times.
We store the parity of each character in a 26-bit mask: bit c is 1 when character c occurs an odd number of times. The characters on a path can be rearranged into a palindrome if and only if this mask has at most one bit set, including the all-zero mask. Equivalently, the bitwise AND of the mask and the mask minus one is 0.
Let pref(u) be the XOR of the masks from the root to node u, inclusive. If w is the lowest common ancestor of u and v, the mask of the path from u to v is
$pref(u) \oplus pref(v) \oplus bit[w].
The portion from the root to w appears once in each prefix and cancels. Node w itself also cancels, so we XOR bit[w] to put it back on the path.
We maintain pref(u) with an Euler tour and an XOR binary indexed tree. During the DFS we record the entry time tin[u] and the exit time tout[u], and add bit[u] at both positions. A prefix query that ends at tin[u] counts each node on the path from the root to u exactly once: its entry time lies inside the prefix and its exit time lies outside. A subtree that has already been fully exited is counted twice and cancels. The tree accumulates from index 1, so query(tin[u] + 1) in the code is pref(u).
When the character of node u changes, let \delta be the XOR of the old and new masks, and XOR \delta at tin[u] and tout[u]. The LCA is precomputed with binary lifting and each query takes O(log n).
The time complexity is O((n + q) log n) and the space complexity is O(n log n), where n is the number of nodes and q$ is the number of operations.
class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.tree = [0] * (n + 1)
def update(self, i: int, delta: int) -> None:
i += 1
while i <= self.n:
self.tree[i] ^= delta
i += i & -i
def query(self, i: int) -> int:
res = 0
while i:
res ^= self.tree[i]
i -= i & -i
return res
class Solution:
def palindromePath(
self, n: int, edges: list[list[int]], s: str, queries: list[str]
) -> list[bool]:
g = [[] for _ in range(n)]
for u, v in edges:
g[u].append(v)
g[v].append(u)
parent = [-1] * n
depth = [0] * n
tin = [0] * n
tout = [0] * n
timer = 0
def dfs(u: int, p: int) -> None:
nonlocal timer
parent[u] = p
tin[u] = timer
timer += 1
for v in g[u]:
if v != p:
depth[v] = depth[u] + 1
dfs(v, u)
tout[u] = timer
dfs(0, -1)
m = n.bit_length()
up = [parent[:]]
up[0][0] = 0
for j in range(1, m):
prev = up[-1]
up.append([prev[prev[i]] for i in range(n)])
def lca(u: int, v: int) -> int:
if depth[u] < depth[v]:
u, v = v, u
d = depth[u] - depth[v]
j = 0
while d:
if d & 1:
u = up[j][u]
d >>= 1
j += 1
if u == v:
return u
for j in range(m - 1, -1, -1):
if up[j][u] != up[j][v]:
u = up[j][u]
v = up[j][v]
return parent[u]
bit = [1 << (ord(c) - ord("a")) for c in s]
tree = BinaryIndexedTree(n)
for u in range(n):
tree.update(tin[u], bit[u])
tree.update(tout[u], bit[u])
ans = []
for q in queries:
op, a, b = q.split()
u = int(a)
if op == "update":
x = bit[u]
y = 1 << (ord(b) - ord("a"))
if x != y:
delta = x ^ y
tree.update(tin[u], delta)
tree.update(tout[u], delta)
bit[u] = y
else:
v = int(b)
w = lca(u, v)
mask = tree.query(tin[u] + 1) ^ tree.query(tin[v] + 1) ^ bit[w]
ans.append(mask & (mask - 1) == 0)
return ansclass BinaryIndexedTree {
int n;
int[] tree;
BinaryIndexedTree(int n) {
this.n = n;
tree = new int[n + 1];
}
void update(int i, int delta) {
for (i++; i <= n; i += i & -i) {
tree[i] ^= delta;
}
}
int query(int i) {
int res = 0;
for (; i > 0; i -= i & -i) {
res ^= tree[i];
}
return res;
}
}
class Solution {
List<Integer>[] g;
int[] parent, depth, tin, tout, bit;
int[][] up;
int timer, m;
public List<Boolean> palindromePath(int n, int[][] edges, String s, String[] queries) {
g = new List[n];
for (int i = 0; i < n; i++) {
g[i] = new ArrayList<>();
}
for (int[] e : edges) {
g[e[0]].add(e[1]);
g[e[1]].add(e[0]);
}
parent = new int[n];
depth = new int[n];
tin = new int[n];
tout = new int[n];
timer = 0;
dfs(0, -1);
m = 32 - Integer.numberOfLeadingZeros(n);
up = new int[m][n];
System.arraycopy(parent, 0, up[0], 0, n);
for (int j = 1; j < m; j++) {
for (int i = 0; i < n; i++) {
up[j][i] = up[j - 1][up[j - 1][i]];
}
}
bit = new int[n];
for (int i = 0; i < n; i++) {
bit[i] = 1 << (s.charAt(i) - 'a');
}
BinaryIndexedTree tree = new BinaryIndexedTree(n);
for (int i = 0; i < n; i++) {
tree.update(tin[i], bit[i]);
tree.update(tout[i], bit[i]);
}
List<Boolean> ans = new ArrayList<>();
for (String q : queries) {
String[] parts = q.split(" ");
int u = Integer.parseInt(parts[1]);
if (parts[0].equals("update")) {
int x = bit[u];
int y = 1 << (parts[2].charAt(0) - 'a');
if (x != y) {
int delta = x ^ y;
tree.update(tin[u], delta);
tree.update(tout[u], delta);
bit[u] = y;
}
} else {
int v = Integer.parseInt(parts[2]);
int w = lca(u, v);
int mask = tree.query(tin[u] + 1) ^ tree.query(tin[v] + 1) ^ bit[w];
ans.add((mask & (mask - 1)) == 0);
}
}
return ans;
}
void dfs(int u, int p) {
parent[u] = p == -1 ? 0 : p;
tin[u] = timer++;
for (int v : g[u]) {
if (v != p) {
depth[v] = depth[u] + 1;
dfs(v, u);
}
}
tout[u] = timer;
}
int lca(int u, int v) {
if (depth[u] < depth[v]) {
int t = u;
u = v;
v = t;
}
int d = depth[u] - depth[v];
for (int j = 0; d > 0; j++, d >>= 1) {
if ((d & 1) != 0) {
u = up[j][u];
}
}
if (u == v) {
return u;
}
for (int j = m - 1; j >= 0; j--) {
if (up[j][u] != up[j][v]) {
u = up[j][u];
v = up[j][v];
}
}
return parent[u];
}
}class BinaryIndexedTree {
int n;
vector<int> tree;
public:
BinaryIndexedTree(int n)
: n(n)
, tree(n + 1) {}
void update(int i, int delta) {
for (++i; i <= n; i += i & -i) {
tree[i] ^= delta;
}
}
int query(int i) {
int res = 0;
for (; i > 0; i -= i & -i) {
res ^= tree[i];
}
return res;
}
};
class Solution {
vector<vector<int>> g, up;
vector<int> parent, depth, tin, tout, bit;
int timer, m;
void dfs(int u, int p) {
parent[u] = p == -1 ? 0 : p;
tin[u] = timer++;
for (int v : g[u]) {
if (v != p) {
depth[v] = depth[u] + 1;
dfs(v, u);
}
}
tout[u] = timer;
}
int lca(int u, int v) {
if (depth[u] < depth[v]) {
swap(u, v);
}
int d = depth[u] - depth[v];
for (int j = 0; d; j++, d >>= 1) {
if (d & 1) {
u = up[j][u];
}
}
if (u == v) {
return u;
}
for (int j = m - 1; j >= 0; j--) {
if (up[j][u] != up[j][v]) {
u = up[j][u];
v = up[j][v];
}
}
return parent[u];
}
public:
vector<bool> palindromePath(int n, vector<vector<int>>& edges, string s, vector<string>& queries) {
g.assign(n, {});
for (auto& e : edges) {
g[e[0]].push_back(e[1]);
g[e[1]].push_back(e[0]);
}
parent.resize(n);
depth.assign(n, 0);
tin.resize(n);
tout.resize(n);
timer = 0;
dfs(0, -1);
m = 32 - __builtin_clz(n);
up.assign(m, vector<int>(n));
up[0] = parent;
for (int j = 1; j < m; j++) {
for (int i = 0; i < n; i++) {
up[j][i] = up[j - 1][up[j - 1][i]];
}
}
bit.resize(n);
for (int i = 0; i < n; i++) {
bit[i] = 1 << (s[i] - 'a');
}
BinaryIndexedTree tree(n);
for (int i = 0; i < n; i++) {
tree.update(tin[i], bit[i]);
tree.update(tout[i], bit[i]);
}
vector<bool> ans;
for (string& q : queries) {
stringstream ss(q);
string op;
int u;
ss >> op >> u;
if (op == "update") {
string c;
ss >> c;
int y = 1 << (c[0] - 'a');
if (bit[u] != y) {
int delta = bit[u] ^ y;
tree.update(tin[u], delta);
tree.update(tout[u], delta);
bit[u] = y;
}
} else {
int v;
ss >> v;
int w = lca(u, v);
int mask = tree.query(tin[u] + 1) ^ tree.query(tin[v] + 1) ^ bit[w];
ans.push_back((mask & (mask - 1)) == 0);
}
}
return ans;
}
};class BinaryIndexedTree {
n: number;
tree: number[];
constructor(n: number) {
this.n = n;
this.tree = Array(n + 1).fill(0);
}
update(i: number, delta: number): void {
for (i++; i <= this.n; i += i & -i) {
this.tree[i] ^= delta;
}
}
query(i: number): number {
let res = 0;
for (; i > 0; i -= i & -i) {
res ^= this.tree[i];
}
return res;
}
}
function palindromePath(n: number, edges: number[][], s: string, queries: string[]): boolean[] {
const g: number[][] = Array.from({ length: n }, () => []);
for (const [u, v] of edges) {
g[u].push(v);
g[v].push(u);
}
const parent = Array(n).fill(0);
const depth = Array(n).fill(0);
const tin = Array(n).fill(0);
const tout = Array(n).fill(0);
let timer = 0;
function dfs(u: number, p: number): void {
parent[u] = p === -1 ? 0 : p;
tin[u] = timer++;
for (const v of g[u]) {
if (v !== p) {
depth[v] = depth[u] + 1;
dfs(v, u);
}
}
tout[u] = timer;
}
dfs(0, -1);
const m = 32 - Math.clz32(n);
const up: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
up[0] = [...parent];
for (let j = 1; j < m; j++) {
for (let i = 0; i < n; i++) {
up[j][i] = up[j - 1][up[j - 1][i]];
}
}
const bit = Array(n).fill(0);
for (let i = 0; i < n; i++) {
bit[i] = 1 << (s.charCodeAt(i) - 97);
}
const tree = new BinaryIndexedTree(n);
for (let i = 0; i < n; i++) {
tree.update(tin[i], bit[i]);
tree.update(tout[i], bit[i]);
}
function lca(u: number, v: number): number {
if (depth[u] < depth[v]) {
[u, v] = [v, u];
}
let d = depth[u] - depth[v];
for (let j = 0; d > 0; j++, d >>= 1) {
if (d & 1) {
u = up[j][u];
}
}
if (u === v) {
return u;
}
for (let j = m - 1; j >= 0; j--) {
if (up[j][u] !== up[j][v]) {
u = up[j][u];
v = up[j][v];
}
}
return parent[u];
}
const ans: boolean[] = [];
for (const q of queries) {
const parts = q.split(' ');
const u = Number(parts[1]);
if (parts[0] === 'update') {
const y = 1 << (parts[2].charCodeAt(0) - 97);
if (bit[u] !== y) {
const delta = bit[u] ^ y;
tree.update(tin[u], delta);
tree.update(tout[u], delta);
bit[u] = y;
}
} else {
const v = Number(parts[2]);
const w = lca(u, v);
const mask = tree.query(tin[u] + 1) ^ tree.query(tin[v] + 1) ^ bit[w];
ans.push((mask & (mask - 1)) === 0);
}
}
return ans;
}type BinaryIndexedTree struct {
n int
tree []int
}
func NewBinaryIndexedTree(n int) *BinaryIndexedTree {
return &BinaryIndexedTree{n: n, tree: make([]int, n+1)}
}
func (b *BinaryIndexedTree) update(i, delta int) {
for i++; i <= b.n; i += i & -i {
b.tree[i] ^= delta
}
}
func (b *BinaryIndexedTree) query(i int) int {
res := 0
for ; i > 0; i -= i & -i {
res ^= b.tree[i]
}
return res
}
func palindromePath(n int, edges [][]int, s string, queries []string) []bool {
g := make([][]int, n)
for _, e := range edges {
u, v := e[0], e[1]
g[u] = append(g[u], v)
g[v] = append(g[v], u)
}
parent := make([]int, n)
depth := make([]int, n)
tin := make([]int, n)
tout := make([]int, n)
timer := 0
var dfs func(int, int)
dfs = func(u, p int) {
if p == -1 {
parent[u] = 0
} else {
parent[u] = p
}
tin[u] = timer
timer++
for _, v := range g[u] {
if v != p {
depth[v] = depth[u] + 1
dfs(v, u)
}
}
tout[u] = timer
}
dfs(0, -1)
m := 1
for (1 << m) <= n {
m++
}
up := make([][]int, m)
up[0] = append([]int(nil), parent...)
for j := 1; j < m; j++ {
up[j] = make([]int, n)
for i := 0; i < n; i++ {
up[j][i] = up[j-1][up[j-1][i]]
}
}
bit := make([]int, n)
for i := 0; i < n; i++ {
bit[i] = 1 << (s[i] - 'a')
}
tree := NewBinaryIndexedTree(n)
for i := 0; i < n; i++ {
tree.update(tin[i], bit[i])
tree.update(tout[i], bit[i])
}
lca := func(u, v int) int {
if depth[u] < depth[v] {
u, v = v, u
}
d := depth[u] - depth[v]
for j := 0; d > 0; j++ {
if d&1 != 0 {
u = up[j][u]
}
d >>= 1
}
if u == v {
return u
}
for j := m - 1; j >= 0; j-- {
if up[j][u] != up[j][v] {
u = up[j][u]
v = up[j][v]
}
}
return parent[u]
}
ans := make([]bool, 0)
for _, q := range queries {
var op string
var u int
fmt.Sscanf(q, "%s %d", &op, &u)
if op == "update" {
var c string
fmt.Sscanf(q, "%s %d %s", &op, &u, &c)
y := 1 << (c[0] - 'a')
if bit[u] != y {
delta := bit[u] ^ y
tree.update(tin[u], delta)
tree.update(tout[u], delta)
bit[u] = y
}
} else {
var v int
fmt.Sscanf(q, "%s %d %d", &op, &u, &v)
w := lca(u, v)
mask := tree.query(tin[u]+1) ^ tree.query(tin[v]+1) ^ bit[w]
ans = append(ans, mask&(mask-1) == 0)
}
}
return ans
}struct BinaryIndexedTree {
n: usize,
tree: Vec<i32>,
}
impl BinaryIndexedTree {
fn new(n: usize) -> Self {
Self {
n,
tree: vec![0; n + 1],
}
}
fn update(&mut self, mut i: usize, delta: i32) {
i += 1;
while i <= self.n {
self.tree[i] ^= delta;
i += i & i.wrapping_neg();
}
}
fn query(&self, mut i: usize) -> i32 {
let mut res = 0;
while i > 0 {
res ^= self.tree[i];
i -= i & i.wrapping_neg();
}
res
}
}
impl Solution {
pub fn palindrome_path(
n: i32,
edges: Vec<Vec<i32>>,
s: String,
queries: Vec<String>,
) -> Vec<bool> {
let n = n as usize;
let mut g = vec![vec![]; n];
for e in &edges {
let u = e[0] as usize;
let v = e[1] as usize;
g[u].push(v);
g[v].push(u);
}
let mut parent = vec![0usize; n];
let mut depth = vec![0usize; n];
let mut tin = vec![0usize; n];
let mut tout = vec![0usize; n];
let mut timer = 0usize;
fn dfs(
u: usize,
p: usize,
g: &[Vec<usize>],
parent: &mut [usize],
depth: &mut [usize],
tin: &mut [usize],
tout: &mut [usize],
timer: &mut usize,
) {
parent[u] = if p == usize::MAX { 0 } else { p };
tin[u] = *timer;
*timer += 1;
for &v in &g[u] {
if v != p {
depth[v] = depth[u] + 1;
dfs(v, u, g, parent, depth, tin, tout, timer);
}
}
tout[u] = *timer;
}
dfs(
0,
usize::MAX,
&g,
&mut parent,
&mut depth,
&mut tin,
&mut tout,
&mut timer,
);
let m = (32 - (n as u32).leading_zeros()) as usize;
let mut up = vec![vec![0usize; n]; m];
up[0].clone_from(&parent);
for j in 1..m {
for i in 0..n {
up[j][i] = up[j - 1][up[j - 1][i]];
}
}
let lca = |mut u: usize, mut v: usize| -> usize {
if depth[u] < depth[v] {
std::mem::swap(&mut u, &mut v);
}
let mut d = depth[u] - depth[v];
let mut j = 0;
while d > 0 {
if d & 1 == 1 {
u = up[j][u];
}
d >>= 1;
j += 1;
}
if u == v {
return u;
}
for j in (0..m).rev() {
if up[j][u] != up[j][v] {
u = up[j][u];
v = up[j][v];
}
}
parent[u]
};
let bytes = s.as_bytes();
let mut bit = vec![0i32; n];
for i in 0..n {
bit[i] = 1 << (bytes[i] - b'a') as u32;
}
let mut tree = BinaryIndexedTree::new(n);
for i in 0..n {
tree.update(tin[i], bit[i]);
tree.update(tout[i], bit[i]);
}
let mut ans = Vec::new();
for q in &queries {
let parts: Vec<&str> = q.split(' ').collect();
let u: usize = parts[1].parse().unwrap();
if parts[0] == "update" {
let x = bit[u];
let y = 1 << (parts[2].as_bytes()[0] - b'a') as u32;
if x != y {
let delta = x ^ y;
tree.update(tin[u], delta);
tree.update(tout[u], delta);
bit[u] = y;
}
} else {
let v: usize = parts[2].parse().unwrap();
let w = lca(u, v);
let mask = tree.query(tin[u] + 1) ^ tree.query(tin[v] + 1) ^ bit[w];
ans.push(mask & (mask - 1) == 0);
}
}
ans
}
}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 |
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.