<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:
    vector<TrieNode*> nodes;
    bool isEnd;

    TrieNode() {
        nodes.resize(2);
        isEnd = false;
    }
};

class Trie {

public:
    TrieNode* root;

    Trie() {
        root = new TrieNode();
        // Write your code here.
    }

    void insert(int num) {
        // Write your code here.

        TrieNode* curr = root;
        for (int i = 31; i >=0; i--) {
            int bit = (num & 1 << i) > 0;
            if (curr->nodes[bit] == NULL) {
                curr->nodes[bit] = new TrieNode();
            }

            curr = curr->nodes[bit];
        }
        curr->isEnd = true;
    }

    int query(int num)
    {
        TrieNode* curr = root;
        int sum= 0;
        if (curr->nodes[0] == NULL && curr->nodes[1] == NULL)
        return -1;

        for (int i = 31; i >=0; i--) {
            int bit = (num & 1 << i) > 0;
            //cout << bit << " ";
            if (curr->nodes[1-bit] != NULL)
            {
               sum+= 1 << i;
               curr = curr->nodes[1 - bit];
            }
            else
            curr = curr->nodes[bit];            
        }

        return sum;
    }

};


class Solution {
public:
    
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<vector<int>> q;
        for (int i = 0; i < queries.size(); i++) {
            auto v = queries[i];
            q.push_back({v[1], v[0], i});
        }

        sort(q.begin(), q.end());
        sort(nums.begin(), nums.end());
        Trie trie;
        vector<int> res(queries.size());
        int x, constraint;
        int j=0;

        for(int i=0; i<q.size(); i++)
        {
            x = q[i][1];
            constraint = q[i][0];
            //cout<<x<<" "<<constraint<<" - ";

            while(j<nums.size() && nums[j]<=constraint)
            {
                //cout<<nums[j]<<" ";
                trie.insert(nums[j]);
                j++;
            }

            int r = trie.query(x);
            res[q[i][2]] = r;
            //cout<<endl;
        }
        return res;
    }
};
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