<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 Node{
        Node* links[2];

        bool containsKey(int x){
            return links[x]!=NULL;
        }

        void putKey(int x , Node* node){
            links[x] = node;
        }

        Node* getKey(int x){
            return links[x];
        }
    };

    class Trie{
        public:
        Node* root;
        Trie(){
            root = new Node();
        }

        void insert(int num){
            Node* temp = root;
            for (int i=30 ; i>=0 ; i--){
                int bit = (num>>i)&1;
                if (!temp->containsKey(bit)){
                    temp->putKey(bit , new Node());
                }
                temp = temp->getKey(bit);
            }
        }

        int compare(int num){
            int ans = 0;
            Node* temp = root;
            for (int i=30 ; i>=0 ; i--){
                int bit = ((num>>i)&1) == 1 ? 0 : 1;
                if (temp->containsKey(bit)){
                    ans |= (1<<i);
                    temp = temp->getKey(bit);
                }
                else{
                    if (bit == 1){
                        temp = temp->getKey(0);
                    }
                    else{
                        temp = temp->getKey(1);
                    }
                }
            }
            return ans;
        }

    };

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin() , nums.end());
        int q = queries.size();
        vector <vector <int>> modified(q);
        for (int i=0 ; i<q ; i++){
            vector <int> temp(3);
            temp[0] = queries[i][0];
            temp[1] = queries[i][1];
            temp[2] = i;
            modified[i] = temp;
        }
        sort(modified.begin() , modified.end() , [&](vector <int> a , vector <int> b){return a[1] < b[1];});

        vector <int> ans(q , -1);
        
        Trie t;
        int ptr1 = 0;
        
        for (int i=0 ; i<q ; i++){
            int x = modified[i][0];
            int m = modified[i][1];
            int idx = modified[i][2];
            while (ptr1<nums.size() && nums[ptr1] <= m){
                t.insert(nums[ptr1]);
                ptr1++;
            }
            if (ptr1 != 0){
                int res = t.compare(x);
                ans[idx] = res;
            }
        }
        return ans;
    }
};
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