<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>

#include <vector>
#include <algorithm>
using namespace std;

struct Node {
    Node* links[2] = {nullptr, nullptr}; // Initialize with nullptr for safety

    bool contains(int bit) { return links[bit] != nullptr; }
    void insert(int bit, Node* node) { links[bit] = node; }
    Node* get(int bit) { return links[bit]; }
};

class Trie {
    Node* root;

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

    int getMax(int num) {
        Node* node = root;
        int ans = 0;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (node->contains(1 - bit)) {
                node = node->get(1 - bit);
                ans |= (1 << i);  // Use |= to accumulate the result
            } else {
                node = node->get(bit);
            }
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int m = queries.size(), n = nums.size();
        vector<int> ans(m);

        // Create a new vector for queries, storing (m, x, original_index)
        vector<vector<int>> offQ;
        for (int i = 0; i < m; i++) {
            vector<int> v;
            v.push_back(queries[i][1]);
            v.push_back(queries[i][0]);
            v.push_back(i);
            offQ.push_back(v);
        }

        // Sort nums and the new offQ vector
        sort(nums.begin(), nums.end());
        sort(offQ.begin(), offQ.end());

        Trie trie;
        int ind = 0;
        for (int i = 0; i < m; i++) {
            int m = offQ[i][0];         // Upper limit from the query
            int x = offQ[i][1];  // The actual query number
            int qInd = offQ[i][2]; // Original index of the query
            
            // Insert numbers into the Trie while they are <= m
            while (ind < n && nums[ind] <= m) {
                trie.insert(nums[ind]);
                ind++;
            }

            // If no number is <= m, the answer is -1
            if (ind == 0)
                ans[qInd] = -1;
            else
                ans[qInd] = trie.getMax(x); // Get the maximum XOR
        }

        return ans;
    }
};

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