<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>

class Trie{
    private:
        vector<Trie*> c;

    public:
    Trie() : c(2,nullptr){}
    void insert(int x){
        Trie* node=this;
        for(int b=30;b>=0;--b){
            int curr=(x>>b)&1;
            if(node->c[curr]==nullptr){
                node->c[curr]=new Trie;
            }
            node=node->c[curr];
        }
    }

    int search(int x){
        int ans=0;
        auto node = this;
        
        for(int b=30;b>=0;--b){
            int curr=(x>>b)&1;

            if(node->c[curr^1]){
                node=node->c[curr^1];
                ans|=(1<<b);
            }
            else if(node->c[curr]){
                node=node->c[curr];
            }
            else return -1;
        }
        return ans;
    }
};

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

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

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

        Trie *trie = new Trie();
        int i=0;
        vector<int> ans(queries.size());
        for(auto &j:q){
            int maxi=j[0],num=j[1],ind=j[2];
            while(i<nums.size() && nums[i]<=maxi){
                trie->insert(nums[i]);
                ++i;
            }

            ans[ind]=trie->search(num);
        }

        return ans;
    }
};
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