<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 trie{
        trie* left = NULL;trie* right = NULL;
        int end = -1;
    };

    trie* root;

    void insert(int x){

        trie* curr = root;
        for(int i=31;i>=0;i--){
            if(x&(1<<i)){
                if(curr->right == NULL)
                curr->right = new trie();
                curr = curr->right;
            }
            else{
                if(curr->left == NULL)
                curr->left = new trie();
                curr = curr->left;
            }
        }
        curr->end = x;
    }

    int findMax(int x){
        //cout<<x<<" ";
        trie* curr = root;
        for(int i=31;i>=0;i--){
            if(curr == NULL)
            return -1;
            if(x&(1<<i)){
                curr = curr->left ? curr->left : curr->right;
            }
            else{
                curr = curr->right ? curr->right : curr->left;
            }
        }
        //cout<<curr->end<<" ";

        return curr->end ^ x;
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        root = new trie();
        sort(nums.begin(), nums.end());
        for(int i=0;i<queries.size();i++){
            queries[i].push_back(i);
        }
        sort(queries.begin(), queries.end(), [](const auto& A, const auto& B){
            return A[1] < B[1];
        });
        int i=0, n=queries.size();
        vector<int> ret(n);
        for(auto query : queries){
            while(i<nums.size() && nums[i] <= query[1]){
                insert(nums[i]);
                i++;
            }

            ret[query[2]] = findMax(query[0]);
        }

        return ret;
    }
};
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