<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 node
{
public:
    node* ch[2];
    bool eos;
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) 
    {
        sort(nums.begin(), nums.end());
        node* root = new node();
        node* curr = root;

        int n = queries.size();
        int m = nums.size();

        vector<int> ans(n, 0);
        vector<pair<pair<int, int>, int>>Q;
        for(int i = 0; i <n; i++)
            Q.push_back({{queries[i][1], queries[i][0]},i});
        sort(Q.begin(), Q.end());

        int j = 0;
        for(int i = 0; i < n; i++)
        {
            int xi = Q[i].first.second;
            int mi = Q[i].first.first;
            int index = Q[i].second;

            //insert required
            curr = root;
            while(j < m && nums[j] <= mi)
            {
                curr = root;
                for(int k = 31; k >= 0; k--)
                {
                    int index = ((nums[j]>>k) & 1);
                    if(curr->ch[index] == NULL)
                        curr->ch[index] = new node();
                    curr = curr->ch[index];
                }
                curr->eos = 1;
                j++;
            }

            //find answer for query
            curr = root;
            int curr_xor = -1;
            for(int k = 31; k >= 0; k--)
            {
                int in = ((xi>>k) & 1);
                if(curr->ch[in] == NULL && curr->ch[1-in] == NULL)
                {
                    break;
                }
                else if(curr->ch[1-in] != NULL)
                {
                    curr_xor = (curr_xor<<1) | 1;
                    curr = curr->ch[1-in];
                }
                else
                {
                    curr_xor = (curr_xor<<1) | 0;
                    curr = curr->ch[in];
                }
            }
            ans[index] = curr_xor;
        }

       
        return ans;
    }
};
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