<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>

struct Node{
    Node* lists[2];
    int flag = false;
    bool Exists(int i){
        return (lists[i]!=NULL);
    }
    void put(int i, Node* node){
        lists[i] = node;
    }
    Node* get(int i){
        return lists[i];
    }
};
class Trie{
    private: Node* root;
    public: 
    Trie(){
        root = new Node();
    }
    void insert(int num ){
        Node* temp = root;
        for(int i=31;i>=0;i--){
            int bit = (num>>i)&1;
            if(!temp->Exists(bit)){
                temp->put(bit, new Node());
            }
            temp = temp->get(bit);
        }
    }

    int checkforMaxxor(int num ){
        Node* temp = root;
        int maxi =0;
        if( temp->lists[0]==NULL && temp->lists[1]==NULL) return -1;
        for(int i=31;i>=0;i--){
            int bit = (num>>i)&1;
            if(temp->Exists(1-bit)){
                maxi = maxi | 1<<i;
                temp = temp->get(1-bit);
            }
            else{
                temp=temp->get(bit);
            }
        }
        return maxi;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Trie ty;
        int n = queries.size();
        vector<vector<int>> indexing;
        for(int i=0;i<n;i++){
            vector<int> temp;
            temp.push_back(queries[i][1]);
            temp.push_back(queries[i][0]);
            temp.push_back(i);
            indexing.push_back(temp);
        }
        sort(indexing.begin(),indexing.end());
        int from =0;
        int to =0;
        vector<int> ans(n);
        sort(nums.begin(),nums.end());
        for(int i=0;i<n;i++){
             to = upper_bound(nums.begin(),nums.end(),indexing[i][0])-nums.begin();
            for(int i=from;i<to;i++){
                ty.insert(nums[i]);
            }
            ans[indexing[i][2]]=ty.checkforMaxxor(indexing[i][1]);
            from = to;
        }
        return ans;
    }
};
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