<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];

    Node()
    {
        links[0]=NULL;
        links[1]=NULL;
    }

    bool containsBit(int bit)
    {
        return links[bit] != NULL;
    }

    void put(int bit,Node* child)
    {
        links[bit]=child;
    }

    Node* get(int bit)
    {
        return links[bit];
    }
};

class Trie
{
    private:
    Node* root;

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

    void insert(int n)
    {
        Node* node=root;

        for(int i=31;i>=0;i--)
        {
            int bit = n>>i & 1 ;

            if(!node->containsBit(bit))
            {
                node->put(bit,new Node);
            }

            node = node->get(bit);
        }
    }

    int maxXor(int n)
    {
        Node* node = root ; 
        int ans=0;
        for(int i=31;i>=0;i--)
        {
            int bit = n>>i & 1;

            if(node->containsBit(!bit))
            {
                 ans = ans | 1<<i;
                 node = node->get(!bit);
            }
            else node= node->get(bit);

        }

        return ans;
    }
};


class Compare {
public:
    bool operator()( vector<int> &l, vector<int> &r) {
        if(l[1]==r[1])
        {
            return l[0] < r[0];
        }
        else return l[1]<r[1];
    }
};
class Solution {
public:

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {


        int n=nums.size();
        int m=queries.size();
        vector<vector<int>> help;

        for(int i=0;i<m;i++)
        {
            help.push_back({queries[i][1],queries[i][0],i});  // so we dont't need custome comaprator
        }

        // i don't know why but coustome comaprator was causing issue on bigger test cases
        // we figured out the reason , custome comparator was giving issue beacuese you did'nt clearly stated
        // return logic you return l[1]<=r[1] which was giving error on large test case
        // so simplify you logic in such case as it is done above
        sort(help.begin(),help.end());
        sort(nums.begin(),nums.end());


        Trie trie;

        vector<int> ans(m,-1);
        int j=0;

        for(int i=0;i<m;i++)
        {
            while(j<n && nums[j] <= help[i][0])
            {
                trie.insert(nums[j]);
                j++;
            }

            if( j>0 )
            ans[help[i][2]] = trie.maxXor(help[i][1]);

            else ans[help[i][2]]=-1;

        }

        
        return ans;

        
    }
};
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