<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>

#include<bits/stdc++.h>

class Node{
    public:
    vector<Node*>v;
    Node(){
        v.resize(2);
    }
};

class Trie{
    public:
    Node*root;
    Trie(){
        root = new Node();
    }
    void insert(const string&s){
        Node*root_copy = root;
        for(int i =0;i<s.size();i++){
            char sk = s[i];
            int val = sk-'0';
            if(root_copy->v[val]){
                root_copy = root_copy->v[val];
            }
            else{
                Node*node = new Node();
                root_copy->v[val] = node;
                root_copy = node;
            }
        }
    }
    int getAns(const string&s){
        int ans = 0;
        Node*root_copy = root;
        for(int i =0;i<s.size();i++){
            char sk = s[i];
            int val = sk-'0';
            char opp_sk = ((s[i]=='0')?'1':'0');
            int opp_val = 1-val;
            if(root_copy->v[opp_val]){
                root_copy = root_copy->v[opp_val];
                ans = ((ans<<1) | 1);
            }
            else{
                root_copy = root_copy->v[val];
                ans = (ans<<1);
            }
        }   
        return ans;
    }
};

class Solution {
public:
    static bool cmp(vector<int>&a,vector<int>&b){
        return a[1]<b[1];
    }
    std::string getBinary(int num){ //O(32) = O(1)
        std::bitset<32> bits(num);
        std::string bitString = bits.to_string();
        return bitString;
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(),nums.end());
        for(int i =0;i<queries.size();i++){
            queries[i].push_back(i);
        }
        sort(queries.begin(),queries.end(),cmp);
        Trie t1;
        vector<int>ans(queries.size());
        int j = 0;
        for(int i =0;i<queries.size();i++){
            int x = queries[i][0];
            int max_val = queries[i][1];
            int index =  queries[i][2];
            if(max_val<nums[0]){
                ans[index] = -1;
            }
            else{
                for(int idx = j;idx<nums.size();idx++){
                    if(nums[idx]<=max_val){
                        t1.insert(getBinary(nums[idx]));
                        j = idx;
                    }
                    else{
                        j = idx;
                        break;
                    }
                }
                ans[index] = t1.getAns(getBinary(x));
            }
        }
        return ans;
    }
};
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