<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 Node{
    Node *l[2];
    void put(int bit,Node *node)
    {
        l[bit]=node;
        return ;
    }
    Node* get(int bit)
    {
        return l[bit];
    }
    bool containsKey(int bit)
    {
        return l[bit]!=NULL;
    }

};
class Trie{
    private:
    Node *root;
    public:
    Trie(){
        root=new Node();
    }
    void insert(int num)
    {
        Node *temp=root;
        for(int i=31;i>=0;i--)
        {
            if(temp==NULL){break;}
            int bit=(num>>i)&1;
            if(!temp->containsKey(bit))
            {
                temp->put(bit,new Node());
            }
            temp=temp->get(bit);
        }
    }
    int maxgive(int num)
    {
        Node *temp=root;
        int ma=0;
        for(int i=31;i>=0;i--)
        {
            int bit=(num>>i)&1;
            if(temp==NULL){break;}
            if(temp->containsKey(1-bit))
            {
                ma=ma|(1<<i);
                temp=temp->get(1-bit);
            }
           else{ temp=temp->get(bit);}
        }
        return ma;
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(),nums.end());
        sort(queries.begin(), queries.end(), [](const vector<int>& a, const vector<int>& b) {
    return a[1] < b[1];
});
         vector<int>answ;
         int k=0;
          Trie *root=new Trie();
        for(int j=0;j<queries.size();j++)
        {
            int ans=0;
           
            int flg=0;
            for(int i=k;i<nums.size()&&nums[i]<=queries[j][1];i++)
            {
                flg=1;
                root->insert(nums[i]);
                k++;
            }
            if(flg==0){ans=-1;}
            else ans=root->maxgive(queries[j][0]);
            answ.push_back(ans);
        }
        
       

        return answ;
    }
};*/
class Node {
public:
    vector<Node *> child;
    Node() {
        child.assign(2, NULL);
    }
};

class Trie {
public:
    Node *root;
    Trie() {
        root = new Node();
    }
    void insert(int n) {
        Node *t = root;
        for (int i = 31; i >= 0; --i) {
            bool bit = ((1 << i) & n);
            if (!t->child[bit])
                t->child[bit] = new Node();
            t = t->child[bit];
        }
    }
    int query(int x) {
        Node *t = root;
        int res = 0;
        for (int i = 31; i >= 0; --i) {
            int bit = (bool)((1 << i) & x);
            if (t->child[bit ^ 1]) {
                res |= (1 << i);
                t = t->child[bit ^ 1];
            } else {
                t = t->child[bit];
            }
        }
        return res;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = (int) queries.size();
        vector<int> ans(n, -1);
        
        vector<int> idx(n);
        iota(idx.begin(), idx.end(), 0);     
        sort(idx.begin(), idx.end(), [&](int i, int j) {
            return queries[i][1] < queries[j][1];
        });
        
        sort(nums.begin(), nums.end());
        Trie t;
        int st = 0;
        for (int i = 0; i < n; ++i) {
            int x = queries[idx[i]][0];
            int m = queries[idx[i]][1];
            while (st < (int) nums.size() && nums[st] <= m) {
                t.insert(nums[st]);
                st++;
            }
            ans[idx[i]] = (st ? t.query(x) : -1);
        }
        return ans;
    }
};
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