<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>

struct Node{
    Node* links[2];
    public:
    bool containsKey(int num){
        return (links[num] != NULL);
    }
    void insert(int num, Node* node){
        links[num] = node;
    }
    Node* get(int num){
        return links[num];
    }
};

class Trie {
    Node* root;
    public:
    Trie(){
        root = new Node();
    }
    public:
    void insert(int num){
        Node*node = root;
        for(int i=31;i>=0;i--){
            int bit = (num>>i) & 1;
            if(!node->containsKey(bit)) node->insert(bit, new Node());
            node = node->get(bit);
        }
    }

    int maxiVal(int num){
        Node* node = root;
        int ans =0;
        for(int i = 31;i>=0;i--){
            int bit = (num>>i) & 1;
                if(node->containsKey(1-bit)) {
                ans = ans | (1<<i);
                node = node->get(1-bit);
                }
                else node = node->get(bit);          
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Trie trie;
        int n = queries.size();
       sort(nums.begin(), nums.end());
        vector<pair<pair<int, int>, int>> arr(n);
        for(int i=0;i<n;i++){
            arr[i].first.first = queries[i][1];
            arr[i].first.second = queries[i][0];
            arr[i].second = i;
        }
        sort(arr.begin(), arr.end());
        vector<int> ans(n);
        int j =0;
         for (int i = 0; i < n; i++) {
            while (j < nums.size() && nums[j] <= arr[i].first.first) {
                trie.insert(nums[j]);
                j++;
            }
            if (j == 0) {  // If no elements were inserted
                ans[arr[i].second] = -1;
            } else {
                ans[arr[i].second] = trie.maxiVal(arr[i].first.second);
            }
        }

        return ans;
    }
};
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