<p>You are given an array <code>nums</code> consisting of non-negative integers. You are also given a <code>queries</code> array, where <code>queries[i] = [x<sub>i</sub>, m<sub>i</sub>]</code>.</p>

<p>The answer to the <code>i<sup>th</sup></code> query is the maximum bitwise <code>XOR</code> value of <code>x<sub>i</sub></code> and any element of <code>nums</code> that does not exceed <code>m<sub>i</sub></code>. In other words, the answer is <code>max(nums[j] XOR x<sub>i</sub>)</code> for all <code>j</code> such that <code>nums[j] &lt;= m<sub>i</sub></code>. If all elements in <code>nums</code> are larger than <code>m<sub>i</sub></code>, then the answer is <code>-1</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em> and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
<strong>Output:</strong> [3,3,7]
<strong>Explanation:</strong>
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
<strong>Output:</strong> [15,-1,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= nums[j], x<sub>i</sub>, m<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

struct Node {
    Node* links[2];
    bool end = false;

    bool contains(int x){
        return links[x] != nullptr;
    }

    Node* get(int x){
        return links[x];
    }

    void put(int x, Node* n){
        links[x] = n;
    }

    void setEnd(){
        end = true;
    }

    bool getEnd(){
        return end;
    }

};


class Trie {
    Node* root;
    public:
        Trie(){
            root = new Node();
        }

        void insert(int num){
            Node* curr = root;

            for(int i = 31; i>=0; i--){
                int t = (num >> i) & 1;
                if(!curr->contains(t)){
                    Node* n = new Node();
                    curr->put(t, n);
                }
                curr = curr->get(t);
            }

            curr->setEnd();
        }

        int getMaxXor(int k){
            int a = 0;
            Node* curr = root;

            for(int i = 31; i>=0; i--){
                int t = (k>>i) & 1;
                t = (t+1)%2;
                if(curr->contains(t)){
                    a = (1 << i) | a;
                    curr = curr->get(t);
                }else if(curr->contains((t+1)%2)){
                    curr = curr->get((t+1)%2);
                }else{
                    return -1;
                }
            }
            return a;
        }

};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(), nums.end());
        vector<int> ans(queries.size());

        vector<vector<int>> new_q;

        for(int i = 0; i<queries.size(); i++){
            vector<int> temp = {queries[i][1], queries[i][0], i};
            new_q.push_back(temp);
        }

        sort(new_q.begin(), new_q.end());

        Trie* t = new Trie();
        int i = 0;
        for(auto q: new_q){
            while(nums[i] <= q[0] && i < nums.size()){
                t->insert(nums[i]);
                i++;
            }
            ans[q[2]] = t->getMaxXor(q[1]);
        }
        
        return ans;
    }
};
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