<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>

struct Trie{
    vector<vector<int>> t;
    vector<int> b;
    int cur = 1;
    Trie() {
        b.assign(2, -1);
        t.push_back(b);
    }
    void insert(int x) {
        int v = 0;
        for(int i=30; i>=0; i--) {
            int bit = (x >> i) & 1;
            if(t[v][bit] == -1) {
                t[v][bit] = cur++;
                t.push_back(b);
            }
            v = t[v][bit];
        }
    }
    int getMax(int x) {
        int v = 0, sum = 0;
        for(int i=30; i>=0; i--) {
            int bit = (x >> i) & 1;
            if(t[v][!bit] != -1) {
                sum+=(1<<i);
                v=t[v][!bit];
            }
            else v = t[v][bit];
            
        }
        return sum;
    }
    
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();

        Trie tr;
        sort(nums.begin(), nums.end());

        vector<array<int, 3>> p;
        int m = queries.size();
        for(int i=0; i<m; i++) {
            p.push_back({queries[i][1], queries[i][0], i});
        }
        sort(p.begin(), p.end());
        int i = 0;
        vector<int>ans(m, -1);
        for(int j=0; j<m; j++) {
            while(i<n && nums[i]<=p[j][0]) {
                tr.insert(nums[i++]);
            }
            if(i==0)continue;
            ans[p[j][2]] = tr.getMax(p[j][1]);
        }
        return ans;
    }
};
422350