<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] = {nullptr, nullptr};

    void put(int bit, Node* node) {
        links[bit] = node;
    }

    Node* get(int bit) {
        return links[bit];
    }

    bool contains(int bit) {
        return links[bit] != nullptr;
    }
};

class Trie {
private:
    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->put(bit, new Node());
            }
            node = node->get(bit);
        }
    }

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

struct Query {
    int x;
    int m;
    int idx;

    bool operator<(const Query& other) const {
        return this->m < other.m;
    }
};

vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
    vector<int> result(queries.size(), -1);
    sort(nums.begin(), nums.end());

    // Create query objects
    vector<Query> queryList;
    for (int i = 0; i < queries.size(); i++) {
        int x = queries[i][0];
        int m = queries[i][1];
        queryList.push_back({x, m, i});
    }

    // Sort queries based on `m`
    sort(queryList.begin(), queryList.end());

    Trie trie;
    int j = 0;

    // Process each query
    for (int i = 0; i < queryList.size(); i++) {
        auto curr = queryList[i];

        // Insert numbers into the Trie that are ≤ `curr.m`
        while (j < nums.size() && nums[j] <= curr.m) {
            trie.insert(nums[j]);
            j++;
        }

        // If there are no numbers ≤ `curr.m`, result will be -1
        if (j == 0) {
            result[curr.idx] = -1;
        } else {
            result[curr.idx] = trie.maxXor(curr.x);
        }
    }

    return result;
}

};
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