{ "data": { "question": { "questionId": "3906", "questionFrontendId": "3590", "categoryTitle": "Algorithms", "boundTopicId": 3705403, "title": "Kth Smallest Path XOR Sum", "titleSlug": "kth-smallest-path-xor-sum", "content": "
You are given an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1. Each node i has an integer value vals[i], and its parent is given by par[i].
The path XOR sum from the root to a node u is defined as the bitwise XOR of all vals[i] for nodes i on the path from the root node to node u, inclusive.
You are given a 2D integer array queries, where queries[j] = [uj, kj]. For each query, find the kjth smallest distinct path XOR sum among all nodes in the subtree rooted at uj. If there are fewer than kj distinct path XOR sums in that subtree, the answer is -1.
Return an integer array where the jth element is the answer to the jth query.
In a rooted tree, the subtree of a node v includes v and all nodes whose path to the root passes through v, that is, v and its descendants.
\n
Example 1:
\n\nInput: par = [-1,0,0], vals = [1,1,1], queries = [[0,1],[0,2],[0,3]]
\n\nOutput: [0,1,-1]
\n\nExplanation:
\n\n
Path XORs:
\n\n11 XOR 1 = 01 XOR 1 = 0Subtree of 0: Subtree rooted at node 0 includes nodes [0, 1, 2] with Path XORs = [1, 0, 0]. The distinct XORs are [0, 1].
Queries:
\n\nqueries[0] = [0, 1]: The 1st smallest distinct path XOR in the subtree of node 0 is 0.queries[1] = [0, 2]: The 2nd smallest distinct path XOR in the subtree of node 0 is 1.queries[2] = [0, 3]: Since there are only two distinct path XORs in this subtree, the answer is -1.Output: [0, 1, -1]
Example 2:
\n\nInput: par = [-1,0,1], vals = [5,2,7], queries = [[0,1],[1,2],[1,3],[2,1]]
\n\nOutput: [0,7,-1,0]
\n\nExplanation:
\n\n
Path XORs:
\n\n55 XOR 2 = 75 XOR 2 XOR 7 = 0Subtrees and Distinct Path XORs:
\n\n[0, 1, 2] with Path XORs = [5, 7, 0]. The distinct XORs are [0, 5, 7].[1, 2] with Path XORs = [7, 0]. The distinct XORs are [0, 7].[2] with Path XOR = [0]. The distinct XORs are [0].Queries:
\n\nqueries[0] = [0, 1]: The 1st smallest distinct path XOR in the subtree of node 0 is 0.queries[1] = [1, 2]: The 2nd smallest distinct path XOR in the subtree of node 1 is 7.queries[2] = [1, 3]: Since there are only two distinct path XORs, the answer is -1.queries[3] = [2, 1]: The 1st smallest distinct path XOR in the subtree of node 2 is 0.Output: [0, 7, -1, 0]
\n
Constraints:
\n\n1 <= n == vals.length <= 5 * 1040 <= vals[i] <= 105par.length == npar[0] == -10 <= par[i] < n for i in [1, n - 1]1 <= queries.length <= 5 * 104queries[j] == [uj, kj]0 <= uj < n1 <= kj <= npar represents a valid tree.给定一棵以节点 0 为根的无向树,带有 n 个节点,按 0 到 n - 1 编号。每个节点 i 有一个整数值 vals[i],并且它的父节点通过 par[i] 给出。
从根节点 0 到节点 u 的 路径异或和 定义为从根节点到节点 u 的路径上所有节点 i 的 vals[i] 的按位异或,包括节点 u。
给定一个 2 维整数数组 queries,其中 queries[j] = [uj, kj]。对于每个查询,找到以 uj 为根的子树的所有节点中,第 kj 小 的 不同 路径异或和。如果子树中 不同 的异或路径和少于 kj,答案为 -1。
返回一个整数数组,其中第 j 个元素是第 j 个查询的答案。
在有根树中,节点 v 的子树包括 v 以及所有经过 v 到达根节点路径上的节点,即 v 及其后代节点。
\n\n
示例 1:
\n\n输入:par = [-1,0,0], vals = [1,1,1], queries = [[0,1],[0,2],[0,3]]
\n\n输出:[0,1,-1]
\n\n解释:
\n\n
路径异或值:
\n\n11 XOR 1 = 01 XOR 1 = 00 的子树:以节点 0 为根的子树包括节点 [0, 1, 2],路径异或值为 [1, 0, 0]。不同的异或值为 [0, 1]。
查询:
\n\nqueries[0] = [0, 1]:节点 0 的子树中第 1 小的不同路径异或值为 0。queries[1] = [0, 2]:节点 0 的子树中第 2 小的不同路径异或值为 1。queries[2] = [0, 3]:由于子树中只有两个不同路径异或值,答案为 -1。输出:[0, 1, -1]
示例 2:
\n\n输入:par = [-1,0,1], vals = [5,2,7], queries = [[0,1],[1,2],[1,3],[2,1]]
\n\n输出:[0,7,-1,0]
\n\n解释:
\n\n
路径异或值:
\n\n55 XOR 2 = 75 XOR 2 XOR 7 = 0子树与不同路径异或值:
\n\n[0, 1, 2],路径异或值为 [5, 7, 0]。不同的异或值为 [0, 5, 7]。[1, 2],路径异或值为 [7, 0]。不同的异或值为 [0, 7]。[2],路径异或值为 [0]。不同的异或值为 [0]。查询:
\n\nqueries[0] = [0, 1]:节点 0 的子树中,第 1 小的不同路径异或值为 0。queries[1] = [1, 2]:节点 1 的子树中,第 2 小的不同路径异或值为 7。queries[2] = [1, 3]:由于子树中只有两个不同路径异或值,答案为 -1。queries[3] = [2, 1]:节点 2 的子树中,第 1 小的不同路径异或值为 0。输出:[0, 7, -1, 0]
\n\n
提示:
\n\n1 <= n == vals.length <= 5 * 1040 <= vals[i] <= 105par.length == npar[0] == -1[1, n - 1] 中的 i,0 <= par[i] < n1 <= queries.length <= 5 * 104queries[j] == [uj, kj]0 <= uj < n1 <= kj <= npar 表示一棵合法的树。u, maintain the set of XOR values along the path from the root to u.",
"Use DSU on tree (small‑to‑large merging) during DFS to efficiently merge each child's set into its parent's set.",
"Store all XOR values in an ordered_set (in Python you can use the sortedcontainers module's SortedList) so you can quickly find the kth smallest XOR in any subtree.",
"At node u, process each query [u, k] by calling find_by_order(k − 1) (C++ PBDS) or indexing sorted_list[k-1] (Python SortedList)."
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