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

class Trie{
private:
    Node*root;
public:

    Trie(){
        root=new Node();
    }
    void insert(int num){
        Node*node=root;
        for(int i=31;i>=0;i--){
            bool k=(num>>i)&1;
            if(!node->containsKey(k)){
                node->put(k,new Node());
            }
            node=node->get(k);
        }
    }

    int getMax(int num){
        Node*node=root;
        int ans=0;
        for(int i=31;i>=0;i--){
            bool k=(num>>i)&1;
            if(node->containsKey(!k)){
                ans=ans | (1<<i); 
                node=node->get(!k);
            }
            else{
                node=node->get(k);
            }
        }
        return ans;
    }
};



class Solution {
public:

    static bool comp(vector<int>&a,vector<int>&b){
        return a[1]<b[1];
    }

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n=queries.size();
        int m=nums.size();
        vector<int>ans(n);
        
        // sort(queries.begin(),queries.end(),comp);
        sort(nums.begin(),nums.end());

        Trie *t=new Trie();

        vector<int> ids(queries.size(), 0);
        iota(ids.begin(), ids.end(), 0); //storing index...0,1,2,3,...
        sort(ids.begin(), ids.end(), [&](int i1, int i2){ 
            return queries[i1][1] < queries[i2][1];
        });   //otherwise u can store the indx of queries and sort and solve the question...see prevoius submission

        int j=0;
        int flag=0;
        for(auto i:ids){ //now ids having indexes in sorted order of Mi`s of queries just use this case and store ans in this ind
            while(j<m && nums[j]<=queries[i][1]){
                flag=1;
                t->insert(nums[j]);
                j++;
            }
            int x=queries[i][0];
            if(flag==1){
                ans[i]=t->getMax(x);
            }
            else{
                ans[i]=-1;
            }
        }
        return ans;
    }
};
200410