<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 Node{
public: 
    Node* bit[2];
    bool flag = false;
    Node(){
        bit[0] = NULL; bit[1] = NULL;
    }
    bool contain(int bit_value){
        return bit[bit_value] == NULL ? false : true;
    }
    void put(int bit_value, Node* node){
        bit[bit_value] = node;
    }
    Node* get(int bit_value){
        return bit[bit_value];
    }
    void setEnd(){
        flag = true;
    }
};

class Solution {
private:
    int find_XOR(int x, Node* node, int place){
        if(place < 0) return 0;

        int bit_value = ((1 << place) & x) == 0 ? 0 : 1;

        if(node -> contain(1 - bit_value)){
            return (1 << place) + find_XOR(x, node -> get(1 - bit_value), place - 1);
        }

        return find_XOR(x, node -> get(bit_value), place - 1);
    }
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node* root = new Node;
        map<vector<int>, int> M;
        vector<vector<int>> dummyQueries; dummyQueries = queries;

        sort(nums.begin(), nums.end());
        sort(queries.begin(), queries.end(), [](const vector<int>& a, const vector<int>& b){
            return a[1] < b[1];
        });
        vector<int> maxXORs(queries.size(), -1);
        int j = 0;

        for(auto q: queries){
            int x = q[0], m = q[1];
            if(M.count({x, m})) continue;

            while(j < nums.size() and nums[j] <= m){
                Node* node = root;

                for(int i = 31; i >= 0; i--){
                    int bit_value = ((1 << i) & nums[j]) == 0 ? 0 : 1;

                    if(!node -> contain(bit_value)) node -> put(bit_value, new Node);
                    node = node -> get(bit_value);
                }
                node -> setEnd();
                j++;
            }

            if(m >= nums[0])  M[{x, m}]= find_XOR(x, root, 31);
            else M[{x, m}] = -1;
        }

        for(int i = 0; i < dummyQueries.size(); i++) maxXORs[i] = M[dummyQueries[i]];

        return maxXORs;
    }
};
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