<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 TrieNode{
    TrieNode* arr[2];
    //bool isEnd{};
};
struct Trie{
    TrieNode* head;
    TrieNode pool[32*100000];
    int poolIndex{};
    Trie(){
        head = &pool[poolIndex++];
        memset(head, 0, sizeof(TrieNode));
    }
    void add(int s){
        TrieNode* curr=head;
        for(int i{31};i>=0;--i){
            int bit = (s>>i)&1;
            if(curr->arr[bit]==nullptr){
                curr->arr[bit]= &pool[poolIndex++];
            memset(curr->arr[bit],0,sizeof(TrieNode));}
            curr=curr->arr[bit];
        }
        return;
    }
    int search(int s){
       //each step... choosing the opposite number possible to maximum xor
       //now need to care the case of not have path at all
       TrieNode* curr = head;
       int current[32];
       bool found = true;
       for(int i{31};i>=0;--i){
        current[31-i]=(s>>i)&1;
       }
       for(int i{};i<32;++i){
            if(curr->arr[!current[i]]){
                curr = curr->arr[!current[i]];
                current[i]=!current[i];
            }else if(curr->arr[current[i]]){
                curr=curr->arr[current[i]];
            }else{
                //not have a path at all
                found=false;
                break;
            }
       }
       if(!found)return -1;
       int ans=0;
       for(int i{};i<32;++i){
        ans=(ans<<1) + current[i];
       }
       //cout<<"biggest xor of "<<s<<" is "<< ans<< " " << (s^ans) <<endl;
       return (s^ans);
    }
};
struct Three{
    int index;
    int num;
    int cap;
};
struct Compare{
    bool operator()(const Three* a, const Three* b){
        return a->cap > b->cap;
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Trie* trie = new Trie();
       //does not exceed? we need to make min heap
       priority_queue<int, vector<int>, greater<int>> add;
       for(int i{};i<nums.size();++i)add.push(nums[i]);
       vector<int> ans(queries.size());
       Three three[queries.size()];
       for(int i{};i<queries.size();++i){
        three[i].index=i;
        three[i].num=queries[i][0];
        three[i].cap=queries[i][1];
       }
       priority_queue<Three*, vector<Three*>, Compare> q;
       for(int i{};i<queries.size();++i)q.push(&three[i]);
       Three* temp;
       while(!q.empty()){
            while(!add.empty() && add.top()<= q.top()->cap){
                trie->add(add.top());add.pop();
            }
            temp = q.top();q.pop();
            ans[temp->index]=trie->search(temp->num);
       }
        return ans;
    }
};
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