<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 Node {
public:
    Node* links[2];

    Node() {
        links[0] = nullptr;
        links[1] = nullptr;
    }

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

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

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

class Trie {
private:
    Node* root;

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

    void insert(int num) {
        int mask = 1 << 31;
        Node* node = root;

        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1; // (num & mask) ? 1 : 0;

            if (!node->containsBit(bit)) {
                node->put(bit, new Node());
            }

            node = node->get(bit);
            mask >>= 1;
        }
    }

    int findMaximumXOR(int num) {
        int mask = 1 << 31, ans = 0;
        Node* node = root;

        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1; //(num & mask) ? 1 : 0;

            if (node->containsBit(!bit)) {
                ans |= (1 << i);
                node = node->get(!bit);
            } else {
                node = node->get(bit);
            }

            mask >>= 1; // Shift the mask to the right
        }
        return ans;
    }
};

class Solution {
private:
    Trie trie;

public:
    static bool comp(pair<vector<int>, int>& a, pair<vector<int>, int>& b) {
        return a.first[1] < b.first[1];
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& qs) {
        int nq = qs.size(), n = nums.size();

        vector<int> ans(nq);
        sort(nums.begin(), nums.end());

        vector<pair<vector<int>, int>> qInd;

        // Populate qInd with pairs of query vectors and their indices
        for (int i = 0; i < nq; i++) {
            qInd.push_back({qs[i], i});
        }

        // Sort qInd using the custom comparator
        sort(qInd.begin(), qInd.end(), comp);

        int alreadyInserted = -1;

        for (auto qrp : qInd) { // qrp = qr pari -> qr, ind
            int ind = qrp.second;
            int x = qrp.first[0];
            int m = qrp.first[1];

            // vector<int> toXorWith;
            // for(int i=0; i<n; i++){
            //     if(nums[i] <= m) toXorWith.push_back(nums[i]);
            //     else break;
            // }

            int upperInd =
                upper_bound(nums.begin(), nums.end(), m) - nums.begin();

            int maxXor = -1;

            if (upperInd > 0) {
                // trie.insert(nums);
                for (int i = alreadyInserted + 1; i < upperInd && i < n; i++) {
                    int ele = nums[i];
                    trie.insert(ele);
                }
                
                alreadyInserted = upperInd-1;

                int xorr = trie.findMaximumXOR(x);
                maxXor = max(maxXor, xorr);
            }

            ans[ind] = maxXor;
        }

        return ans;
    }
};
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