<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>

class Trie{
public:
    Trie *zero, *one;
    int minimum;

    Trie(){
        zero = NULL;
        one = NULL;
        minimum = INT_MAX;
    }

    void insert(int num){
        Trie *ptr = this;
        ptr->minimum = min(ptr->minimum, num);
        for(int i = 31; i >= 0; i--){
            if(num&(1 << i)){
                if(!ptr->one){
                    ptr->one = new Trie;
                }
                ptr = ptr->one;
            }
            else{
                if(!ptr->zero){
                    ptr->zero = new Trie;
                }
                ptr = ptr->zero;
            }
            ptr->minimum = min(ptr->minimum, num);
        }
    }
};

class Solution {
public:
    int query(Trie *trie, int x, int m){
        Trie *ptr = trie;
        int ans = 0;
        // kahin pe m ka bit 1 tha but humein 0 ki jrurat thi isliye zero ke saath proceed kiya
        // flag = true hone ke baad m ka bit kahin pe 0 hai tab bhi wahan pe 1 le skte hain
        bool flag = false;
        ptr = trie;
        for(int i = 31; i >= 0; i--){
            if(ptr->minimum > m){
                return -1;
            }
            if(flag || m&(1 << i)){
                if(x&(1 << i)){ // x ka i-th bit 1 hai -> 0 ki taraf jayengein
                    if(ptr->zero && ptr->zero->minimum <= m){
                        ans += (1 << i);
                        ptr = ptr->zero;
                        flag = true;
                    }
                    else{
                        ptr = ptr->one;
                    }
                }
                else{ // x ka i-th bit 0 hai -> 1 ki taraf jayengein
                    if(ptr->one && ptr->one->minimum <= m){
                        ans += (1 << i);
                        ptr = ptr->one;
                    }
                    else{
                        ptr = ptr->zero;
                        flag = true;
                    }
                }
            }
            else{
                ans += x&(1 << i);
                ptr = ptr->zero;
            }
        }
        return ans;
    }

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size(), q = queries.size();
        Trie trie;
        for(int i = 0; i < n; i++){
            trie.insert(nums[i]);
        }
        vector<int> ans(q);
        for(int i = 0; i < q; i++){
            ans[i] = query(&trie, queries[i][0], queries[i][1]);
        }
        return ans;
    }
};
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