<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];
    bool isEnd;
    bool containsKey(char c) {
        return links[c - '0'] != NULL;
    }
    void addKey(char c, Node* node) {
        links[c - '0'] = node;
    }
    Node* getKey(char c) {
        return links[c - '0'];
    }
    void setEnd() {
        isEnd = true;
    }
};
class Trie {
public:
    Node* node;
    Trie() {
        node = new Node();
    }
    
    void insert(string word) {
        Node* root = node;
        for(char c: word) {
            if(!root -> containsKey(c)) {
                root -> addKey(c, new Node());
            }
            root = root -> getKey(c);
        }
        root -> setEnd();
    }
    
    bool search(string word) {
        Node* root = node;
        for(char c: word) {
            if(!root -> containsKey(c)) {
                return false;
            }
            root = root -> getKey(c);
        }
        return root -> isEnd;
    }
    
    bool startsWith(string prefix) {
        Node* root = node;
        for(char c: prefix) {
            if(!root -> containsKey(c)) {
                return false;
            }
            root = root -> getKey(c);
        }
        return true;
    }
    string getMax(string num) {
        Node* root = node;
        string s = "";
        for(char c: num) {
            char ch = c == '0' ? '1': '0';
            if(!root -> containsKey(ch)) {
                s += '0';
                root = root -> getKey(c);
            }
            else {
                s += '1';
                root = root -> getKey(ch);
            }
        }
        return s;
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        int q = queries.size();
        for(int i = 0; i < q; i++) {
            queries[i].push_back(i);
        }
        sort(queries.begin(), queries.end(), [&](const auto &x, const auto &y) {
           return x[1] < y[1]; 
        });
        vector<int> ans(q, -1);
        Trie* trie = new Trie();
        auto toBinary = [&](int n) {
            string s = "";
            for(int i = 0; i < 32; i++) {
                s += (n % 2 ? '1' : '0');
                n /= 2;
            }
            reverse(s.begin(), s.end());
            return s;
        };
        auto toDecimal = [&](string s) {
            int ans = 0, bit = 0;
            reverse(s.begin(), s.end());
            for(char c: s) {
                ans += (1LL << bit++) * (c - '0');
            }
            return ans;
        };
        int i = 0;
        sort(nums.begin(), nums.end());
        for(auto &qr: queries) {
            int x = qr[0], m = qr[1], index = qr[2];
            if(nums[0] > m) {
                continue;
            }
            while(i < n && nums[i] <= m) {
                trie -> insert(toBinary(nums[i]));
                i++;
            }
            ans[index] = toDecimal(trie -> getMax(toBinary(x)));
        }
        return ans;
    }
};
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