<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 Solution {
public:
    struct TrieNode{
        TrieNode* child[2];
    };
    void insert(TrieNode* node, int num){
        for(int i=31; i>=0; i--){
            int j = (num>>i)&1;
            if(!node->child[j]) node->child[j] = new TrieNode();
            node = node->child[j];
        }
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        TrieNode* root = new TrieNode();
        int n = nums.size(), q = queries.size();
        vector<vector<int>> newQueries;
        vector<int> ans(q, -1);
        for(int i=0; i<q; i++) newQueries.push_back({queries[i][1], queries[i][0], i});
        sort(newQueries.begin(), newQueries.end());
        sort(nums.begin(), nums.end());
        int idx = 0;
        for(auto v : newQueries){
            int m = v[0], x = v[1], ii = v[2];
            if(m<nums[0]) continue;
            while(nums[idx] <= m and idx<n){
                TrieNode* node = root;
                insert(node, nums[idx]);
                idx++;
            }
            int XOR = 0;
            TrieNode* node = root;
            for(int i=31; i>=0; i--){
                int j = (x>>i)&1;
                int notJ = !j;
                if(node->child[notJ]){
                    XOR|= (1<<i);
                    j=notJ;
                }
                node = node->child[j];
            }
            ans[ii] = max(ans[ii], XOR);
        }
        return ans;        
    }
};
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