<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 Solution {
public:
int ans=0;
static bool comp (vector<int>&a,vector<int>&b){
	return a[1]<b[1];
}
struct trie{
trie *next[2];
};

trie *getNode(){
	trie * curr=new trie();
	curr->next[0]=NULL;
	curr->next[1]=NULL;
	return curr;
}

void insert(trie* root, int num){
vector<int>arr(32,0);
int idx=0;
while(num){
	if(num & 1)
	 arr[32-idx-1]=1;
	 num>>=1;
	 idx++;
}
trie * curr=root;
for(int i=0;i<32;i++){
	int x=arr[i];
	int y=x^1;
    
		if(curr->next[x])
		curr=curr->next[x];
                else {
                curr->next[x] = getNode();
                curr = curr->next[x];
                }
}
}

void check(trie* root ,int num){
	vector<int>arr(32,0);
int idx=0;
while(num){
	if(num & 1)
	 arr[32-idx-1]=1;
	 num>>=1;
	 idx++;
}
trie * curr=root;
int val=0;
for (int i = 0; i < 32; i++) {
         int x = arr[i];
         int y = x ^ 1;

         if (curr->next[y])
 {
         curr = curr->next[y];
         val |= (1 << (32 - i - 1));
 }
                else {
                curr = curr->next[x];
                }
}
ans=max(ans,val);
}
    vector<int> maximizeXor(vector<int>& arr, vector<vector<int>>& queries) {
          sort(arr.begin(),arr.end());
   for(int i=0;i<queries.size();i++)
   queries[i].push_back(i);
   sort(queries.begin(),queries.end(),comp);
   
  int i=0,j=0;
  int n=arr.size();
  int qSize=queries.size();
  vector<int>ansF(qSize,-1);
 
 trie * root=getNode();

  while(j<qSize){
   while (i < n && arr[i] <= queries[j][1]) {
         insert(root,arr[i]);
                i++;
   }
        // we are at an index greater -> either i==n or arr[i]>queries[j][1]
	ans=0;
	if(i>0){
		check(root,queries[j][0]);
		ansF[queries[j][2]]=ans;
	}
j++;
  }
    return ansF;  
    }
};
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