<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 Trie{

    public:
        Trie *child[2];

        Trie()
        {
            for(int i=0;i<2;i++)
                child[i] = NULL;
        }
        

};




string makebits(int num) {
    bitset<32> bits(num);
    string str = bits.to_string();
    return str;
}

int findMax(int num,Trie* root)
{
    string str = makebits(num);
    int res = 0;
    Trie* node = root;
    for(auto ch : str)
    {

        char rev = (ch == '0')?'1':'0';

        if(node->child[rev-'0'] == NULL)
        {
            node = node->child[!(rev-'0')];
            res = res<<1;
        }else
        {
            node = node->child[rev - '0'];
            res = res<<1 | 1;
        }
    }

    return res;
    
}

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        
        
        vector<vector<int>> qr;
        int n = queries.size();

        for(int i=0;i<n;i++)
        {
            qr.push_back({queries[i][1],queries[i][0],i});
        }

        vector<int> ans(n,-1);

        sort(nums.begin(),nums.end());
        sort(qr.begin(),qr.end());
        int index = 0;

        Trie *root = new Trie();

        for(int i=0;i<n;i++)
        {
            int lim = qr[i][0];
            int x = qr[i][1];
            int ind = qr[i][2];

            while(index < nums.size() && nums[index] <= lim)
            {

                string str = makebits(nums[index]);

                Trie *node = root;
                for(int i=0;i<str.size();i++)
                {
                    if(node->child[str[i]-'0']==NULL)
                    {
                        node->child[str[i]-'0'] = new Trie();

                    }
                    node = node->child[str[i]-'0'];
                }
                index++;
                
            }

            if (index != 0) {
                int res = findMax(x, root);
                ans[ind] = res;
            }
        }

        return ans;


        
    }
};
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