3841. Palindromic Path Queries in a Tree

Difficulty: Hard

Topics: Principal, Array, String, Divide and Conquer, Bit Manipulation, Tree, Depth-First Search, Binary Indexed Tree, Segment Tree, Biweekly Contest 176

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 node uᵢ to c. Formally, update s[uᵢ] = c.
"query uᵢ vᵢ": Determine whether the string formed by the characters on the unique path from uᵢ to vᵢ (inclusive) can be rearranged into a palindrome¹.

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": 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]. Example 3:
Input: n = 1, edges = [], s = "a", queries = ["query 0 0"]
Output: [true]

Example 4:

Input: n = 2, edges = [[0,1]], s = "aa", queries = ["query 0 1"]
Output: [true]

Example 5:

Input: n = 2, edges = [[0,1]], s = "ab", queries = ["query 0 1"]
Output: [false]

Example 6:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[3,4]], s = "abcba", queries = ["query 0 4"]
Output: [true]

Example 7:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[3,4]], s = "abcde", queries = ["query 0 4"]
Output: [false]

Example 8:

Input: n = 4, edges = [[0,1],[1,2],[2,3]], s = "aaaa", queries = ["update 1 a","query 0 3"]
Output: [true]

Example 9:

Input: n = 4, edges = [[0,1],[1,2],[2,3]], s = "aaaa", queries = ["update 1 b","update 2 b","query 0 3"]
Output: [true]

Example 10:

Input: n = 4, edges = [[0,1],[1,2],[2,3]], s = "aaaa", queries = ["update 1 b","query 0 3"]
Output: [true]

Example 11:

Input: n = 5, edges = [[0,1],[0,2],[0,3],[0,4]], s = "abcde", queries = ["query 1 2","query 3 4","update 0 a","query 1 2"]
Output: [false,false,false]

Example 12:

Input: n = 6, edges = [[0,1],[1,2],[2,3],[3,4],[4,5]], s = "aabbcc", queries = ["query 0 5","update 3 a","query 0 5"]
Output: [true,false]

Example 13:

Input: n = 3, edges = [[0,1],[1,2]], s = "aba", queries = ["query 0 0","query 1 1","query 0 2"]
Output: [true,true,true]

Example 14:

Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]], s = "abcdefg", queries = ["update 3 a","update 5 a","update 1 a","query 0 6"]
Output: [false]

Example 15:

Input: n = 6, edges = [[0,1],[0,2],[1,3],[1,4],[2,5]], s = "xyzzyx", queries = ["query 3 4","query 3 5","update 0 x","query 3 5"]
Output: [true,false,true]

Constraints:

1 <= n == s.length <= 5 * 10⁴
edges.length == n - 1
edges[i] = [uᵢ, vᵢ]
0 <= uᵢ, vᵢ <= 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 uᵢ c" or
- queries[i] = "query uᵢ vᵢ"
- 0 <= uᵢ, vᵢ <= n - 1
- c is a lowercase English letter.

Hint:

Use heavy light decomposition to break each path query into O(log n) segments.
Represent characters as a 26-bit mask and maintain segment data with XOR. A path can form a palindrome if the resulting mask has at most one bit set.

Solution:

We solve the dynamic palindromic-path query problem on a tree using an Euler-tour representation combined with a Fenwick tree for range-XOR/point-query updates, binary lifting for LCA, and 26-bit integer masks to represent character parities. Each character is encoded as a single bit; a string is rearrangeable into a palindrome iff at most one bit in the XOR mask of its characters is set. By observing that the path u → v mask equals prefix(u) XOR prefix(v) XOR mask[LCA(u,v)], and that the prefix mask at a node changes exactly when a node in the subtree of an updated node is affected, we reduce each update to a subtree-range XOR and each query to two point queries plus an LCA lookup.

Approach

Character encoding as bitmask: Map each letter c to 1 << (ord(c) - 97). The parity of character counts along any path is the XOR of these masks.
Palindrome criterion: A multiset of characters can be rearranged into a palindrome iff at most one character has odd frequency, i.e., the mask has popcount(mask) <= 1, equivalently (mask & (mask - 1)) == 0.
Root-to-node prefix: Define pref[x] = XOR of masks on path root → x. Then for any u, v: pathMask(u,v) = pref[u] XOR pref[v] XOR mask[LCA(u,v)].
Euler tour: Flatten the tree so each node's subtree is a contiguous interval [tin[u], tout[u]]. This lets us model "all descendants of u" as a range.
Fenwick tree with range-XOR / point-query: When node u's mask changes by delta, pref[x] changes iff x is in u's subtree. So apply delta as a range XOR over [tin[u], tout[u]]; a point query pref[x] is then the accumulated XOR at tin[x].
Binary lifting LCA: Precompute up[k][v] for O(log n) ancestor jumps, enabling O(log n) LCA queries.
Offline processing: Parse each query string; if update, compute delta = oldMask XOR newMask (skip if 0) and apply a subtree range XOR; if query, compute pathMask and append the palindrome test result.

Let's implement this solution in PHP: 3841. Palindromic Path Queries in a Tree

<?php /** * @param Integer $n * @param Integer[][] $edges * @param String $s * @param String[] $queries * @return Boolean[] */ function palindromePath(int $n, array $edges, string $s, array $queries): array { // Build adjacency list. $adj = array_fill(0, $n, []); foreach ($edges as $e) { $u = $e[0]; $v = $e[1]; $adj[$u][] = $v; $adj[$v][] = $u; } // Euler tour: tin[u]..tout[u] is the subtree interval of u. $parent = array_fill(0, $n, 0); $depth = array_fill(0, $n, 0); $tin = array_fill(0, $n, 0); $tout = array_fill(0, $n, 0); $timer = 0; $stack = [[0, 0, 0]]; // [node, parent, state], state 0 = enter, 1 = exit $parent[0] = 0; while (!empty($stack)) { $item = array_pop($stack); $u = $item[0]; $p = $item[1]; $state = $item[2]; if ($state === 0) { $parent[$u] = $p; $tin[$u] = $timer++; $stack[] = [$u, $p, 1]; foreach ($adj[$u] as $v) { if ($v === $p) continue; $depth[$v] = $depth[$u] + 1; $parent[$v] = $u; $stack[] = [$v, $u, 0]; } } else { $tout[$u] = $timer - 1; } } // Binary lifting table for LCA. $LOG = 1; while ((1 << $LOG) <= $n) { $LOG++; } $up = array_fill(0, $LOG, array_fill(0, $n, 0)); for ($i = 0; $i < $n; $i++) { $up[0][$i] = $parent[$i]; } for ($k = 1; $k < $LOG; $k++) { for ($i = 0; $i < $n; $i++) { $up[$k][$i] = $up[$k - 1][$up[$k - 1][$i]]; } } // Character masks. $mask = array_fill(0, $n, 0); for ($i = 0; $i < $n; $i++) { $mask[$i] = 1 << (ord($s[$i]) - 97); } /* * Fenwick tree for range XOR update and point query. * * If a node u changes by delta, every root-to-x prefix XOR changes * exactly when x is in the subtree of u. So we XOR delta over * [tin[u], tout[u]]. */ $N = $n + 1; $bit = array_fill(0, $N + 1, 0); $add = function($i, $val) use (&$bit, $N) { while ($i <= $N) { $bit[$i] ^= $val; $i += $i & -$i; } }; $rangeXor = function($l, $r, $val) use ($add) { if ($val === 0) return; $add($l + 1, $val); $add($r + 2, $val); }; $pointQuery = function($idx0) use (&$bit) { $i = $idx0 + 1; $res = 0; while ($i > 0) { $res ^= $bit[$i]; $i -= $i & -$i; } return $res; }; // Initialize prefix XORs by applying each node's initial mask // to its whole subtree. for ($i = 0; $i < $n; $i++) { $rangeXor($tin[$i], $tout[$i], $mask[$i]); } // LCA query. $lca = function($a, $b) use (&$up, $depth, $LOG) { if ($depth[$a] < $depth[$b]) { $tmp = $a; $a = $b; $b = $tmp; } $diff = $depth[$a] - $depth[$b]; for ($k = 0; $k < $LOG; $k++) { if ($diff & (1 << $k)) { $a = $up[$k][$a]; } } if ($a === $b) { return $a; } for ($k = $LOG - 1; $k >= 0; $k--) { if ($up[$k][$a] !== $up[$k][$b]) { $a = $up[$k][$a]; $b = $up[$k][$b]; } } return $up[0][$a]; }; $answer = []; foreach ($queries as $query) { $parts = explode(' ', $query); if ($parts[0] === 'update') { $u = (int)$parts[1]; $newMask = 1 << (ord($parts[2]) - 97); $delta = $mask[$u] ^ $newMask; if ($delta !== 0) { $rangeXor($tin[$u], $tout[$u], $delta); $mask[$u] = $newMask; } } else { $u = (int)$parts[1]; $v = (int)$parts[2]; $w = $lca($u, $v); // XOR parity mask on path u -> v. $pathMask = $pointQuery($tin[$u]) ^ $pointQuery($tin[$v]) ^ $mask[$w]; // Palindrome possible iff at most one bit is set. $answer[] = (($pathMask & ($pathMask - 1)) === 0); } } return $answer; } // Test cases echo palindromePath(3, [[0,1],[1,2]], "aac", ["query 0 2","update 1 b","query 0 2"]) . "\n"; // Output: [true,false] echo palindromePath(4, [[0,1],[0,2],[0,3]], "abca", ["query 1 2","update 0 b","query 2 3","update 3 a","query 1 3"]) . "\n"; // Output: [false,false,true] echo palindromePath(1, [], "a", ["query 0 0"]) . "\n"; // Output: [true] echo palindromePath(2, [[0,1]], "aa", ["query 0 1"]) . "\n"; // Output: [true] echo palindromePath(2, [[0,1]], "ab", ["query 0 1"]) . "\n"; // Output: [false] echo palindromePath(5, [[0,1],[1,2],[2,3],[3,4]], "abcba", ["query 0 4"]) . "\n"; // Output: [true] echo palindromePath(5, [[0,1],[1,2],[2,3],[3,4]], "abcde", ["query 0 4"]) . "\n"; // Output: [false] echo palindromePath(4, [[0,1],[1,2],[2,3]], "aaaa", ["update 1 a","query 0 3"]) . "\n"; // Output: [true] echo palindromePath(4, [[0,1],[1,2],[2,3]], "aaaa", ["update 1 b","update 2 b","query 0 3"]) . "\n"; // Output: [true] echo palindromePath(4, [[0,1],[1,2],[2,3]], "aaaa", ["update 1 b","query 0 3"]) . "\n"; // Output: [true] echo palindromePath(5, [[0,1],[0,2],[0,3],[0,4]], "abcde", ["query 1 2","query 3 4","update 0 a","query 1 2"]) . "\n"; // Output: [false,false,false] echo palindromePath(6, [[0,1],[1,2],[2,3],[3,4],[4,5]], "aabbcc", ["query 0 5","update 3 a","query 0 5"]) . "\n"; // Output: [true,false] echo palindromePath(3, [[0,1],[1,2]], "aba", ["query 0 0","query 1 1","query 0 2"]) . "\n"; // Output: [true,true,true] echo palindromePath(7, [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]], "abcdefg", ["update 3 a","update 5 a","update 1 a","query 0 6"]) . "\n"; // Output: [false] echo palindromePath(6, [[0,1],[0,2],[1,3],[1,4],[2,5]], "xyzzyx", ["query 3 4","query 3 5","update 0 x","query 3 5"]) . "\n"; // Output: [true,false,true] ?>

Explanation:

Tree preprocessing:
- Build an adjacency list from edges.
- Perform an iterative DFS producing tin[u], tout[u] (subtree interval), depth[u], and parent[u]. Euler order guarantees tin[u] < tin[v] < tout[v] < tout[u] when u is an ancestor of v.
LCA preprocessing:
- Fill up[0][v] = parent[v] and iteratively up[k][v] = up[k-1][up[k-1][v]] for k = 1..LOG-1.
- LCA procedure: lift the deeper node, then lift both together to the highest differing ancestors, returning the parent of that point.
Fenwick initialization:
- Start with an all-zero bit array; for every node u, perform rangeXor(tin[u], tout[u], mask[u]). After this, the point query at tin[x] equals pref[x] — the XOR of masks from root to x.
Update handling (update u c):
- Compute newMask = 1 << (c - 'a'), delta = mask[u] XOR newMask.
- If delta != 0, do rangeXor(tin[u], tout[u], delta) and set mask[u] = newMask.
- Range XOR is done with two Fenwick point updates (+1 and +2 on the exclusive end) since Fenwick supports prefix XOR for point queries.
Query handling (query u v):
- Find w = LCA(u, v).
- pathMask = pointQuery(tin[u]) XOR pointQuery(tin[v]) XOR mask[w].
- Push ((pathMask & (pathMask - 1)) == 0) into the answer.
Why it works:
- pointQuery(tin[u]) XOR pointQuery(tin[v]) cancels the common root-to-LCA prefix, leaving the XOR of the two LCA → u and LCA → v branches excluding the LCA. XOR-ing mask[w] re-inserts the LCA exactly once, giving the inclusive path mask.