<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:
class node
{
    public:
    node* children[2];
    int freq;
    int terminal;
    node()
    {
        int i;
        for(i=0;i<2;i++)
        {
            children[i] = NULL;
        }
        freq = 0;
        terminal = 0;

    }
    bool contains(int ch)
    {
        if(children[ch]!=NULL && children[ch]->freq!=0)
        return true;
        else
        return false;
    }
    node* get(int ch)
    {
        return children[ch];
    }
    void put(int ch, node*n)
    {
        children[ch] = n;
    }
    void markend()
    {
        terminal++;
    }
};

class Trie{

    public:
    node* root;
    Trie() {
        root = new node();
    }
   
    void insert(vector<int> &word){
        int i , n = word.size();
        node* curr = root;
        curr->freq++;
        for(i=0;i<n;i++)
        {
            if(!curr->contains(word[i]))
            {
                curr->children[word[i]] = new node();
            }
            curr = curr->children[word[i]];
            curr->freq++;
        }
        curr->markend();
    }

};
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(),nums.end());
        int i,j, n = nums.size();
        Trie* t = new Trie();
        vector<vector<int>> v(n,vector<int>(32));
       
        for(i=0;i<n;i++)
        {
            int temp = nums[i];
            int c=0;
            while(temp>0)
            {
                if(temp%2)
                v[i][31-c]=1;
                temp/=2;
                c++;
            }
        }
        int m = queries.size();
         vector<int> ans(m);
        for(i=0;i<m;i++)
        {
            swap(queries[i][0],queries[i][1]);
            queries[i].push_back(i);
        }
        sort(queries.begin(),queries.end());
        for(i=0;i<m;i++)
        {
            cout<<queries[i][0]<<" "<<queries[i][1]<<endl;
        }
        int k=0;
        for(i=0;i<m;i++)
        {
            while(k<n && nums[k]<=queries[i][0])
            {
                t->insert(v[k]);
                k++;
            }
            if(k==0)
            {
                    ans[queries[i][2]]=-1;
                    continue;
            }
            vector<int> dig(32,0);
            int temp = queries[i][1];
            int c=0;
            while(temp>0)
            {
                if(temp%2)
                dig[31-c]=1;
                temp/=2;
                c++;
            }
             node* curr = t->root;
             int maxi=0;
            for(j=0;j<32;j++)
            {
                
                if(curr->children[abs(1-dig[j])])
                {
                    maxi += pow(2,31-j);
                    curr = curr->children[abs(1-dig[j])];
                }
                else
                {
                    curr = curr->children[dig[j]];
                }
            }
            ans[queries[i][2]]=maxi;
        }
        return ans;
    }
};
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