<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 Trie{
    vector<vector<int>> trie;
    int nc;
    public:
    Trie(){
        nc=1;
        trie.push_back(vector<int>(2,-1));
    }
    void insert(int num){
        int currNode=0;
        for(int i=31;i>=0;i--){
            int bit=(num>>i)&1;
            if(trie[currNode][bit]==-1){
                trie.push_back(vector<int>(2,-1));
                trie[currNode][bit]=nc;
                nc++;
            }
            currNode=trie[currNode][bit];
        }
    }
    int maxXor(int num){
        int maxi = 0;
        int curr = 0;
        for (int i = 31; i >= 0; i--) {
            int cbit = (num >> i) & 1;
            int obit = 1 - cbit;
            if (trie[curr][obit] != -1) {
                maxi |= (1 << i);
                curr = trie[curr][obit];
            } else {
                curr = trie[curr][cbit];
            }
        }
        return maxi;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& arr, vector<vector<int>>& queries) {
        vector<int> ans(queries.size(), 0); 
        vector<pair<int, pair<int,int>>> offlineQueries;
        sort(arr.begin(), arr.end()); 
        
        int index = 0;
        
        for(auto &it: queries) {
            offlineQueries.push_back({it[1],{it[0], index++}}); 
        }
        
        sort(offlineQueries.begin(), offlineQueries.end());
        int i = 0; 
        int n = arr.size(); 
        
        Trie trie; 
        
        for(auto &it : offlineQueries) {
            while(i < n && arr[i] <= it.first) {
                trie.insert(arr[i]); 
                i++; 
            }
            if(i != 0) 
                ans[it.second.second] = trie.maxXor(it.second.first); 
            else 
                ans[it.second.second] = -1; 
        }
        
        return ans; 
    }
};
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