<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 *links[2];

    bool containsKey(int bit){
        return (links[bit]!=NULL);
    }

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

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

class Trie{
    private:
    Node* root;

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

    void insert(int num){
        Node *node = root;

        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(!node->containsKey(bit)){
                node->put(bit, new Node());
            }
            node = node->get(bit);
        }
    }

    int getMax(int num){
        Node *node = root;

        int maxNum=0;
        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(node->containsKey(1 - bit)){
                maxNum |= (1<<i);
                node = node->get(1- bit);
            }
            else{
                node = node->get(bit);
            }
        }
        return maxNum;
    }

};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<int> ans(queries.size(), 0);

        vector<pair<int, pair<int, int>>> offlineQueries;

        for(int i=0; i<queries.size(); i++){
            offlineQueries.push_back({queries[i][1], {queries[i][0], i}});
        }
        
        sort(nums.begin(), nums.end());
        sort(offlineQueries.begin(), offlineQueries.end());

        int i=0;
        Trie trie;
        int n = nums.size();

        for(auto &it : offlineQueries){
            int mi = it.first; 
            int xi = it.second.first;
            int ind = it.second.second;

            while(i<n && nums[i]<=mi){
                trie.insert(nums[i++]);
            }

            //no element inserted in trie
            if(i==0){
                ans[ind] = -1;
            }
            else{
                ans[ind] = trie.getMax(xi);
            }

        }

        return ans;

    }
};
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