Hard
Maximum XOR of Two Non-Overlapping Subtrees — C++
Full explanation · Time O(nlogr) · Space O(n)
// Time: O(nlogr), r is sum(values)
// Space: O(n)
// iterative dfs, trie, greedy
class Solution {
private:
class Trie {
public:
Trie(int bit_length)
: bit_length_(bit_length)
, nodes_(1) {}
void insert(int64_t num) {
int idx = 0;
for (int i = bit_length_ - 1; i >= 0; --i) {
int64_t curr = (num >> i) & 1;
if (!nodes_[idx][curr]) {
nodes_.emplace_back();
nodes_[idx][curr] = size(nodes_) - 1;
}
idx = nodes_[idx][curr];
}
}
int64_t query(int64_t num) {
if (size(nodes_) == 1) {
return -1;
}
int64_t result = 0, idx = 0;
for (int i = bit_length_ - 1; i >= 0; --i) {
int64_t curr = (num >> i) & 1;
if (nodes_[idx][1 ^ curr]) {
idx = nodes_[idx][1 ^ curr];
result |= 1ll << i;
} else {
idx = nodes_[idx][curr];
}
}
return result;
}
private:
const int bit_length_;
vector<array<int, 2>> nodes_;
};
public:
long long maxXor(int n, vector<vector<int>>& edges, vector<int>& values) {
vector<vector<int>> adj(size(values));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
const auto& iter_dfs = [&]() {
vector<int64_t> lookup(size(values));
vector<tuple<int, int, int>> stk = {{1, 0, -1}};
while (!empty(stk)) {
const auto [step, u, p] = stk.back(); stk.pop_back();
if (step == 1) {
stk.emplace_back(2, u, p);
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
stk.emplace_back(1, v, u);
}
} else if (step == 2) {
lookup[u] = values[u];
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
lookup[u] += lookup[v];
}
}
}
return lookup;
};
const auto& lookup = iter_dfs();
Trie trie(bit_length(lookup[0]));
const auto& iter_dfs2 = [&]() {
using RET = int64_t;
RET result{0};
vector<tuple<int, int, int, shared_ptr<RET>, RET *>> stk = {{1, 0, -1, nullptr, &result}};
while (!empty(stk)) {
const auto [step, u, p, new_ret, ret] = stk.back(); stk.pop_back();
if (step == 1) {
*ret = max(*ret, trie.query(lookup[u]));
stk.emplace_back(3, u, p, new_ret, ret);
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
const auto& new_ret = make_shared<RET>(0);
stk.emplace_back(2, v, u, new_ret, ret);
stk.emplace_back(1, v, u, nullptr, new_ret.get());
}
} else if (step == 2) {
*ret = max(*ret, *new_ret);
} else if (step == 3) {
trie.insert(lookup[u]);
}
}
return result;
};
return iter_dfs2();
}
private:
int bit_length(int64_t x) {
return x != 0 ? 64 - __builtin_clzll(x) : 1;
}
};
// Time: O(nlogr), r is sum(values)
// Space: O(n)
// dfs, trie, greedy
class Solution2 {
private:
class Trie {
public:
Trie(int bit_length)
: bit_length_(bit_length)
, nodes_(1) {}
void insert(int64_t num) {
int idx = 0;
for (int i = bit_length_ - 1; i >= 0; --i) {
int64_t curr = (num >> i) & 1;
if (!nodes_[idx][curr]) {
nodes_.emplace_back();
nodes_[idx][curr] = size(nodes_) - 1;
}
idx = nodes_[idx][curr];
}
}
int64_t query(int64_t num) {
if (size(nodes_) == 1) {
return -1;
}
int64_t result = 0, idx = 0;
for (int i = bit_length_ - 1; i >= 0; --i) {
int64_t curr = (num >> i) & 1;
if (nodes_[idx][1 ^ curr]) {
idx = nodes_[idx][1 ^ curr];
result |= 1ll << i;
} else {
idx = nodes_[idx][curr];
}
}
return result;
}
private:
const int bit_length_;
vector<array<int, 2>> nodes_;
};
public:
long long maxXor(int n, vector<vector<int>>& edges, vector<int>& values) {
vector<vector<int>> adj(size(values));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1]);
adj[e[1]].emplace_back(e[0]);
}
vector<int64_t> lookup(size(values));
const function<int64_t(int, int)> dfs = [&](int u, int p) {
lookup[u] = values[u];
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
lookup[u] += dfs(v, u);
}
return lookup[u];
};
dfs(0, -1);
Trie trie(bit_length(lookup[0]));
const function<int64_t(int, int)> dfs2 = [&](int u, int p) {
int64_t result = max(trie.query(lookup[u]), static_cast<int64_t>(0));
for (const auto& v : adj[u]) {
if (v == p) {
continue;
}
result = max(result, dfs2(v, u));
}
trie.insert(lookup[u]);
return result;
};
return dfs2(0, -1);
}
private:
int bit_length(int64_t x) {
return x != 0 ? 64 - __builtin_clzll(x) : 1;
}
};