<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{
    TrieNode* bit[2];
    public:
    bool containsKey(int num){
        return bit[num]!=NULL;
    }
    TrieNode* getKey(int num){
        return bit[num];
    }
    void putKey(int num,TrieNode* node){
        bit[num] = node;
    }
};
class Trie{
    public:
    TrieNode* root;
    Trie(){
        root = new TrieNode();
    }
    void insert(int num){
        TrieNode* node = root;
        for(int i=31;i>=0;i--){
            int bit = (num>>i)&1;
            if(!node->containsKey(bit)){
                node->putKey(bit,new TrieNode());
            }
            node = node->getKey(bit);
        }
    }
    int getMaxXor(int num){
        TrieNode* node = root;
        int maxi = 0;
        for(int i=31;i>=0;i--){
            int bit = (num>>i)&1;
            if(node->containsKey(1-bit)){
                maxi = maxi | (1<<i);
                node = node->getKey(1-bit);
            }else{
                node = node->getKey(bit);
            }
        }
        return maxi;
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> arr;
        for(int i=0;i<queries.size();i++){
            arr.push_back({queries[i][1],queries[i][0],i});
        }
        sort(arr.begin(),arr.end());
        int ind = 0;
        Trie T;
        int exor = 0;
        vector<int>ans(queries.size());
        for(int i=0;i<queries.size();i++){
            int val = arr[i][0];
            int num = arr[i][1];
            int pos = arr[i][2];
            while(ind<nums.size() && nums[ind]<=val){
                T.insert(nums[ind++]);
            }
            if(ind==0) exor = -1;
            else exor = T.getMaxXor(num);
            ans[pos]=exor;
        }
        return ans;
    }
};
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