<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
{
    Node* bits[2];
    public:
    Node()
    {
        for(int i = 0; i < 2; i++)
        {
            bits[i] = NULL;
        }
    }

    bool containsBit(int bit)
    {
        return (bits[bit] != NULL);
    }

    Node* get(int bit)
    {
        return bits[bit];
    }

    void put(int bit, Node *node)
    {
        bits[bit] = node;
    }
};

class Trie
{
    Node *root;
    public:
    Trie()
    {
        root = new Node();
    }

    void insert(int num)
    {
        Node *curr = root;
        for(int i = 31; i >= 0; i--)
        {
            int bit = (num >> i) & 1;
            if(!curr->containsBit(bit))
            {
                curr->put(bit, new Node());
            }
            curr = curr->get(bit);
        }
    }

    int getmax(int num)
    {
        Node *curr = root;
        int ans = 0;
        for(int i = 31; i >= 0; i--)
        {
            int bit = (num >> i) & 1;
            if(curr->containsBit(1 - bit))
            {
                bit = 1 - bit;
                ans = ans | (1 << i);
            }
            curr = curr->get(bit);
        }
        return ans;
    }
};

bool comp(pair<int, pair<int, int>> &p1, pair<int, pair<int, int>> &p2)
{
    return p1.second.second < p2.second.second;
}
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size(), m = queries.size();
        Trie *allnum = new Trie();
        sort(nums.begin(), nums.end());
        vector<pair<int, pair<int, int>>> squeries(m);
        for(int i = 0; i < m; i++)
        {
            squeries[i] = {i, {queries[i][0], queries[i][1]}};
        }
        vector<int> ans(m);
        sort(squeries.begin(), squeries.end(), comp);
        int pn = 0;
        for(int i = 0; i < m; i++)
        {
            while(pn < n && nums[pn] <= squeries[i].second.second)
            {
                allnum->insert(nums[pn]);
                pn++;
            }
            if(pn != 0)
                ans[squeries[i].first] = allnum->getmax(squeries[i].second.first);
            else
                ans[squeries[i].first] = -1;
        }
        return ans;
    }
};
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