<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] = {NULL};
    bool ispre(int bit){
        return links[bit] != NULL;
    }
    void insertbit(int bit , Node* node){
        links[bit] = node;
    }

};

class Trie{
    private: Node* root;
    public: Trie(){
                root = new Node();
            }
    public:
        void insert(int num){
            Node* node = root;
            for(int i=31;i>=0;i--){
                int bit = (num>>i)&1;
                if(!node->ispre(bit)){
                    node->insertbit(bit, new Node());
                }
                node = node->links[bit];
            }
        }
        int getmax(int num){
            int maxi = 0;
            Node* node = root;
            for(int i=31;i>=0;i--){
                int bit = (num>>i)&1;
                if(node->ispre(1-bit)){
                    maxi = maxi | (1<<(i));
                    node= node->links[1-bit];
                }
                else{
                    node = node->links[bit];
                }
            }
            return maxi;
        }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        //here we use the concept of offline queries
        Trie* trie = new Trie();
        vector<vector<int>> qq;
        int i = 0;
        for(auto x: q){
            qq.push_back({x[1], x[0], i++});
        }
        sort(qq.begin(), qq.end());
        sort(nums.begin(), nums.end());
        for(auto x: qq){
            for(auto y: x)cout<<y<<" ";
            cout<<endl;
        }
        vector<int> ans(q.size());
        int idx = 0;
        for(auto x: qq){
            auto fidx = x[2];
            auto maxi = x[0];
            auto num = x[1];
            while(idx < nums.size() && nums[idx] <= maxi){
                trie->insert(nums[idx]);
                idx++;
            }
            if(idx == 0){
                ans[fidx] = -1;
                continue;
            }
            ans[fidx] = trie->getmax(num);
        }
        return ans;
    }
};
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