<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>

bool isset(int n, int j) { return n & (1 << j); }
void setbit(int& n, int j) { n |= (1 << j); }

struct Node {
    Node* links[2]; 

    int mini0 = 2e9, mini1 = 2e9;

    bool isPresent(int bit) { return links[bit] != NULL; }
    void put(int bit, Node* node) { links[bit] = node; }
    Node* get(int bit) { return links[bit]; }
    void storeMin0(int a) { mini0 = min(mini0, a); }
    void storeMin1(int a) { mini1 = min(mini1, a); }
    int getMin0() { return mini0; }
    int getMin1() { return mini1; }
};

class Trie {
private:
    Node* root;

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

    void insert(int n) {
        Node* currnode = root;
        for (int i = 30; i >= 0; --i) {
            int bit = 0;
            if (isset(n, i)) {
                bit = 1;
                currnode->storeMin1(n);
            } else {
                currnode->storeMin0(n);
            }

            if (!(currnode->isPresent(bit))) {
                currnode->put(bit, new Node());
            }
            currnode = currnode->get(bit);
        }
    }

    int findMax(int a, int mx) {
        Node* currnode = root;
        int ans = 0;
        for (int i = 30; i >= 0; --i) {
            int mini0 = currnode->getMin0(), mini1 = currnode->getMin1();
            if (isset(a, i) && currnode->isPresent(0) && mini0 <= mx) {
                currnode = currnode->get(0);
                setbit(ans, i); // xor=1
                continue;
            }
            if (!(isset(a, i)) && currnode->isPresent(1) && mini1 <= mx) {
                currnode = currnode->get(1);
                setbit(ans, i); // xor=1
                continue;
            }
            if (!(isset(a, i)) && currnode->isPresent(0) && mini0 <= mx) {
                currnode = currnode->get(0);
                continue;
            }
            if (isset(a, i) && currnode->isPresent(1) && mini1 <= mx) {
                currnode = currnode->get(1);
                continue;
            }
            else
                return -1;
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Trie ds;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            ds.insert(nums[i]);
        }
        vector<int> ans;
        int q = queries.size();
        for (int i = 0; i < q; ++i) {
            int a = queries[i][0], mx = queries[i][1];
            int ca = ds.findMax(a, mx);
            ans.push_back(ca);
        }
        return ans;
    }
};
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