Hard
Maximum XOR With an Element From Array — C++
Full explanation · Time O(nlogn + mlogm + nlogk + mlogk) · Space O(nlogk)
// Time: O(nlogn + mlogm + nlogk + mlogk), k is max(max(nums), max(xi))
// Space: O(nlogk)
class Trie {
public:
Trie(int bit_length)
: bit_length_(bit_length)
, nodes_(1) {}
void insert(int num) {
int idx = 0;
for (int i = bit_length_; i >= 0; --i) {
int curr = (num >> i) & 1;
if (!nodes_[idx][curr]) {
nodes_.emplace_back();
nodes_[idx][curr] = size(nodes_) - 1;
}
idx = nodes_[idx][curr];
}
}
int query(int num) {
if (size(nodes_) == 1) {
return -1;
}
int result = 0, idx = 0;
for (int i = bit_length_; i >= 0; --i) {
int curr = (num >> i) & 1;
if (nodes_[idx][1 ^ curr]) {
idx = nodes_[idx][1 ^ curr];
result |= 1 << i;
} else {
idx = nodes_[idx][curr];
}
}
return result;
}
private:
const int bit_length_;
vector<array<int, 2>> nodes_;
};
class Solution {
public:
vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
sort(begin(nums), end(nums));
int max_val = max(nums.back(),
(*max_element(cbegin(queries), cend(queries),
[](const auto& a, const auto& b) {
return a[0] < b[0];
}))[0]);
for (int i = 0; i < size(queries); ++i) {
queries[i].insert(begin(queries[i]), i);
}
sort(begin(queries), end(queries),
[](const auto& a, const auto& b) {
return a[2] < b[2];
});
vector<int> result(size(queries));
Trie trie(bit_length(max_val));
int j = 0;
for (const auto& q : queries) {
int i = q[0], x = q[1], m = q[2];
for (; j < size(nums) && nums[j] <= m; ++j) {
trie.insert(nums[j]);
}
result[i] = trie.query(x);
}
return result;
}
private:
int bit_length(int x) {
return x != 0 ? 32 - __builtin_clz(x) : 1;
}
};