<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* bit[2];
};

class Trie{

public:
    Node* root;

    Trie(){
        root = new Node();
    }

    void insert(int x){
        vector<int>binary(32,0);
        int i = 0;
        
        while(x){
            int val = x & 1;
            binary[31 - i] = val;
            i++;
            x = x >> 1;
        }
        
        Node* cur = root;
        for(int i = 0 ; i < 32 ; i ++){
            if(cur->bit[binary[i]] == NULL){
                cur->bit[binary[i]] = new Node();
            }
            cur = cur->bit[binary[i]];
        }
    }

    int getMaxXOR(int val){
        vector<int>binary(32,0);
        int i = 0;
        
        while(val){
            int num = val & 1;
            binary[31 - i] = num;
            i++;
            val = val >> 1;
        }
        
        Node* cur = root;
        int ans = 0;
        for(int i = 0 ; i < 32; i++){
            if(cur->bit[1 - binary[i]] != NULL){
                ans+=(1<<(31-i));
                cur = cur->bit[1 - binary[i]];
            }
            else{
                cur = cur->bit[binary[i]];
            }
        }

        return ans;
    }
};

class Solution {
public:

    static bool comp(pair<vector<int>,int>&a, pair<vector<int>, int>&b){
        if(a.first[1] == b.first[1]){
            return a.first[0] < b.first[0];
        }
        return a.first[1] < b.first[1];
    }

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<pair<vector<int>,int>>v;
        sort(nums.begin(), nums.end());

        for(int i = 0 ; i < queries.size();i++){
            v.push_back({queries[i],i});
        }

        sort(v.begin(), v.end(), comp);
        vector<int>ans(queries.size(),-1);

        Trie* trie = new Trie();
        int ptr = 0;

        for(int i = 0 ; i < v.size() ; i++){
            while(ptr < nums.size() && nums[ptr] <= v[i].first[1]){
                trie->insert(nums[ptr]);
                ptr++;
            }

            if(ptr == 0){
                ans[v[i].second] = -1;
                continue;
            }

            int val = trie->getMaxXOR(v[i].first[0]);
            ans[v[i].second] = val;
        }

        return ans;
    }
};
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