Hard
Kth Smallest Path XOR Sum — C++
Full explanation · Time O(n * (logn)^2 + qlogn) · Space O(n + q)
// Time: O(n * (logn)^2 + qlogn)
// Space: O(n + q)
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
// iterative dfs, small-to-large merging, ordered set
class Solution {
public:
vector<int> kthSmallest(vector<int>& par, vector<int>& vals, vector<vector<int>>& queries) {
vector<vector<int>> adj(size(par));
for (int u = 0; u < size(par); ++u) {
const auto& p = par[u];
if (p != -1) {
adj[p].emplace_back(u);
}
}
vector<vector<int>> lookup(size(adj));
for (int i = 0; i < size(queries); ++i) {
lookup[queries[i][0]].emplace_back(i);
}
const auto& iter_dfs = [&]() {
using ordered_set = tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;
vector<ordered_set> os(size(adj));
vector<int> idxs(size(adj));
iota(begin(idxs), end(idxs), 0);
const auto& small_to_large_merge = [&](auto& i, auto& j) { // Total Time: O(n * (logn)^2)
if (size(os[i]) < size(os[j])) {
swap(i, j); // swapping indices is much faster than swapping ordered sets
}
for (const auto& x : os[j]) { // each node is merged at most O(logn) times
os[i].insert(x); // each add costs O(logn)
}
};
vector<int> result(size(queries), -1);
vector<tuple<int, int, int>> stk = {{1, 0, 0}};
while (!empty(stk)) {
auto [step, u, curr] = stk.back(); stk.pop_back();
if (step == 1) {
curr ^= vals[u];
os[idxs[u]].insert(curr);
stk.emplace_back(2, u, curr);
for (const auto& v : adj[u]) {
stk.emplace_back(1, v, curr);
}
} else if (step == 2) {
for (const auto& v : adj[u]) {
small_to_large_merge(idxs[u], idxs[v]);
}
for (const auto& i : lookup[u]) { // Total Time: O(qlogn)
if (queries[i][1] - 1 < size(os[idxs[u]])) {
result[i] = *(os[idxs[u]].find_by_order(queries[i][1] - 1));
}
}
}
}
return result;
};
return iter_dfs();
}
};
// Time: O(n * (logn)^2 + qlogn)
// Space: O(n + q)
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
// dfs, small-to-large merging, ordered set
class Solution2 {
public:
vector<int> kthSmallest(vector<int>& par, vector<int>& vals, vector<vector<int>>& queries) {
vector<vector<int>> adj(size(par));
for (int u = 0; u < size(par); ++u) {
const auto& p = par[u];
if (p != -1) {
adj[p].emplace_back(u);
}
}
vector<vector<int>> lookup(size(adj));
for (int i = 0; i < size(queries); ++i) {
lookup[queries[i][0]].emplace_back(i);
}
const auto& small_to_large_merge = [&](auto& ptr1, auto& ptr2) { // Total Time: O(n * (logn)^2)
if (size(*ptr1) < size(*ptr2)) {
swap(ptr1, ptr2); // swapping ptrs is much faster than swapping ordered sets
}
for (const auto& x : *ptr2) { // each node is merged at most O(logn) times
ptr1->insert(x); // each add costs O(logn)
}
};
vector<int> result(size(queries), -1);
using ordered_set = tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;
const function<unique_ptr<ordered_set> (int, int)> dfs = [&](int u, int curr) {
curr ^= vals[u];
auto ptr = make_unique<ordered_set>();
ptr->insert(curr);
for (const auto& v : adj[u]) {
auto new_ptr = dfs(v, curr);
small_to_large_merge(ptr, new_ptr);
}
for (const auto& i : lookup[u]) { // Total Time: O(qlogn)
if (queries[i][1] - 1 < size(*ptr)) {
result[i] = *(ptr->find_by_order(queries[i][1] - 1));
}
}
return ptr;
};
dfs(0, 0);
return result;
}
};