<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 TrieNode{
    public:
        TrieNode * one,*zero;
        TrieNode(){
          one=NULL;
          zero=NULL;
      }
};
void insert(TrieNode *  t,int num){
        vector<int> v(32,0);
        int c=0;
        while(num!=0){
               int r=num%2;
               v[c++]+=r;
               num/=2;
        }
        TrieNode *temp = t;
        c=31;
         while(c>=0){
            if (v[c]==0){
                  if (!temp->zero){
                      temp->zero=new TrieNode();
                  }
                   temp=temp->zero;
            }
            else {
                    if (!temp->one){
                      temp->one=new TrieNode();
                   }
                    temp=temp->one; 
            }
            c--;
         }
}
  int findMax(TrieNode *t, int num) {
    TrieNode *temp = t;
    int ans = 0;
    
    for (int i = 31; i >= 0; i--) {
        int bit = (num >> i) & 1;
        
        if (bit == 0) {
            if (temp->one != NULL) {
                ans |= (1 << i); // This adds 2^i to ans
                temp = temp->one;
            } else if (temp->zero != NULL) {
                temp = temp->zero;
            } else {
                return -1; // No valid path exists
            }
        } else {
            if (temp->zero != NULL) {
                ans |= (1 << i); // This adds 2^i to ans
                temp = temp->zero;
            } else if (temp->one != NULL) {
                temp = temp->one;
            } else {
                return -1; // No valid path exists
            }
        }
    }
    
    return ans;
}
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
          int i,j,k,n=nums.size(),q=queries.size();
           TrieNode * t=new TrieNode();
          vector<int> ans(q);
          vector<vector<int>> v;
          for (i=0;i<q;i++){
             v.push_back({queries[i][1],i});
          }
           sort(v.begin(),v.end());
           sort(nums.begin(),nums.end());
            int x=0;
           for (auto &i:v){
                      int ind=i[1];
                      int maxValue=i[0];
                      int value=queries[ind][0];
                    while(x<n && nums[x]<=maxValue){
                          insert(t,nums[x]);
                          x++;
                    }  
                     cout<<x<<" "<<endl;
                    ans[ind]=findMax(t,value);
           }
            return ans;
    }
};
292885