<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 TNode {
public:
    TNode* child[2];
    TNode(){
        this->child[0] = NULL;
        this->child[1] = NULL;
    }

};

class Trie{
public:
    TNode* root;
    Trie(){
        this->root = new TNode();
    }

    void insert(int& n){
        TNode* cur = root;
        bitset<32> bs(n);

        for(int i = 31; i >= 0; i--){
            if(!cur->child[bs[i]]){
                cur->child[bs[i]] = new TNode();
            }
            cur = cur->child[bs[i]];
        }
    }
    int maxXOR(int& n){
        TNode* cur = root;
        bitset<32> bs(n);
        int res = 0;
        for(int i = 31; i >= 0; i--){
            if(cur->child[!bs[i]]) {
                res += (1 << i);
                cur = cur->child[!bs[i]];
            }
            else cur = cur->child[bs[i]];
        }
        return res;
    }

};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        Trie trie;

        int q = queries.size();
        map<int, map<int, set<int>>> qToI;
        for(int i = 0; i < q; i++){
            int num = queries[i][0];
            int m = queries[i][1];
            qToI[num][m].insert(i);
        }
        sort(queries.begin(), queries.end(),[&](vector<int>& a, vector<int>& b){
            return a[1] < b[1];
        });
        

        vector<int> ans(q, -1);
        int j = 0;
        for(int i = 0; i < q; i++){
            int num = queries[i][0];
            int m = queries[i][1];
            while(j < n and nums[j] <= m){
                trie.insert(nums[j]);
                j++;
            }
            if(j > 0 and m >= nums[j - 1]){
                ans[*qToI[num][m].begin()] = trie.maxXOR(num);
                qToI[num][m].erase(*qToI[num][m].begin());
            }
        }
        return ans;
    }
};
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