<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 <bits/stdc++.h>
using namespace std;
using ints = int;
using doublee = double;
using vi = vector<ints>;
using pi = pair<ints, ints>;
using vvi = vector<vector<ints>>;
using vpi = vector<pi>;
#define endl       "\n"
#define all(arr) arr.begin(), arr.end()

struct node {
    ints prefixCount, wordCount;
    node* children[2];

    node(ints prefixCount) {
        this->prefixCount = prefixCount;
        this->wordCount = 0;
        for (ints i = 0; i < 2; i++)
            children[i] = nullptr;
    }
};

class Trie {
public:
    node* root;

    Trie() {
        root = new node(0);
    }

    void insert(ints num) {
        node* ptr = this->root;
        for (ints i = 30; i >= 0; i--) {
            bool ch = num & (1ll << i);
            if (ptr->children[ch] == nullptr) {
                node* newNode = new node(1);
                ptr->children[ch] = newNode;
            } else {
                ptr->children[ch]->prefixCount++;
            }
            ptr = ptr->children[ch];
        }
        ptr->wordCount++;
    }

    ints maxXor(ints x) {
        ints ans = 0;
        node* ptr = this->root;
        for (ints i = 30; i >= 0; i--) {
            bool ch = x & (1ll << i);
            if (ptr->children[!ch]) {
                ans += (1ll << i);
                ptr = ptr->children[!ch];
            } else ptr = ptr->children[ch];
        }
        return ans;
    }
};

class Solution {
public:
    static bool comp(const vi& a, const vi& b) {
        return a[1] < b[1];
    }

    vi maximizeXor(vi& arr, vvi& q) {
        ints n = arr.size(), m = q.size();
        sort(all(arr));

        for (ints i = 0; i < m; i++) {
            q[i].push_back(i);
        }
        sort(all(q), comp);

        Trie* trie = new Trie();
        vi uniq, sol(m);
        if (!arr.empty()) {
            uniq.push_back(arr[0]);
            for (ints i = 1; i < n; i++) {
                if (arr[i] != arr[i - 1]) {
                    uniq.push_back(arr[i]);
                }
            }
        }
        ints n1 = uniq.size();
        ints curr = -1;
        for (ints i = 0; i < m; i++) {
            ints x = q[i][0], y = q[i][1];
            ints ind = upper_bound(all(uniq), y) - uniq.begin();
            if (ind == 0) {
                sol[q[i][2]] = -1;
                continue;
            }
            for (ints j = curr + 1; j < ind; j++) trie->insert(uniq[j]);
            curr = ind - 1;
            sol[q[i][2]] = trie->maxXor(x);
        }
        return sol;
    }
};

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