<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 Trie {
public:
    struct TrieNode {
        TrieNode* child[2];
        bool isend;

        TrieNode(){
            child[0] = nullptr;
            child[1] = nullptr;
            isend = false;
        }
    };
    

    TrieNode* root;

    Trie() {
        root = new TrieNode();
    }

    void insert_bit(int num) {
        TrieNode* curr = root;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (curr->child[bit] == nullptr) {
                curr->child[bit] = new TrieNode();
            }
            curr = curr->child[bit];
        }
        curr->isend = true;
    }

    int maxxor(int num) {
        TrieNode* curr = root;
      
        int res = 0;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            int oppositeBit = 1 - bit;
            if(curr==nullptr) return -1;
            if (curr->child[oppositeBit] != nullptr) {
                res = (res << 1) | 1;
                curr = curr->child[oppositeBit];
            } else {
                res = (res << 1);
                curr = curr->child[bit];
            }
        }
        return res;
    }
};


class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin() , nums.end());
        int n=nums.size() , m=queries.size();
        int i=0;
        int j=0;
        vector<vector<int>> q;
        int k=0;
        for(auto it:queries){
            q.push_back({it[1] , it[0], k});
           
            k++;
        }
        sort(q.begin() , q.end());

        vector<int> ans(m ,0);
        Trie word;
      
        while(j<m){
            if(i<n && nums[i]<=q[j][0]){
                word.insert_bit(nums[i]);
                i++;
            }
            else{
                ans[q[j][2]]=word.maxxor(q[j][1]);
                j++;
            }

        }

        return ans;
    }
};
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