<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 TrieNode {
public:

    vector<TrieNode*> next;
    TrieNode() {

        next = vector<TrieNode*>(2, nullptr);
    }
};
class Trie{

public:
TrieNode* baseptr=nullptr;
	Trie(){
		baseptr=new TrieNode();

	}
	void insert(int val){
		TrieNode* trieptr=baseptr;
		for (int i = 29; i >= 0; i--) {
			int bit = (val >> i) & 1;
			if (trieptr->next[bit] == nullptr) {
				trieptr->next[bit] = new TrieNode();
			}
			trieptr = trieptr->next[bit];
		}

	}
	int find_max(int val){
		int ans=0;
		TrieNode* trieptr=baseptr;
		for(int i=29;i>=0;i--){

			int bit=(val>>i)&1;
			if(bit == 0) {
				if(trieptr->next[1] != nullptr) {
					ans=ans|(1<<i);
					trieptr = trieptr->next[1];
				} else {
					trieptr = trieptr->next[0];
				}
			} else {
				if(trieptr->next[0] != nullptr) {
					ans=ans|(1<<i);
					trieptr = trieptr->next[0];
				} else {
					trieptr = trieptr->next[1];
				}
			}
					
		}
        return ans;

	}
};
class Solution {
public:



    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
	vector<int> ans(queries.size(), -1);
	for(int i=0;i<queries.size();i++){
		queries[i].push_back(i);
	}
    sort(queries.begin(), queries.end(), 
         [](const vector<int>& a, const vector<int>& b) {
             return a[1] < b[1];
         });
        
    
	sort(nums.begin(),nums.end());

	int j=0;
	Trie trie;

	for(int i=0;i<queries.size();i++){
        while(j<nums.size() && nums[j]<=queries[i][1]){

			trie.insert(nums[j]);

			j++;
		}
		if(j>0){
			int ans1 = trie.find_max(queries[i][0]);
            ans[queries[i][2]] = ans1 ;
			
		}

	}
	return ans;
    }
};
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