<p>Given an integer array <code>nums</code>, return <em>the maximum result of </em><code>nums[i] XOR nums[j]</code>, where <code>0 &lt;= i &lt;= j &lt; n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,10,5,25,2,8]
<strong>Output:</strong> 28
<strong>Explanation:</strong> The maximum result is 5 XOR 25 = 28.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [14,70,53,83,49,91,36,80,92,51,66,70]
<strong>Output:</strong> 127
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Node{
    public : 
        Node* link[2];
        Node() = default;
        void set(int x,Node* n){
            link[x] = n;
        }

        Node* next(int x){
            return link[x];
        }

        bool exists(int x){
            return link[x]!=NULL;
        }
};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        ios_base::sync_with_stdio(NULL),cin.tie(0),cout.tie(0);
        Node* root = new Node();
        for(int i=0;i<nums.size();i++){
            int a = nums[i];
            Node* node =root;
            for(int j=31;j>=0;j--){
                int x = a&(1<<j)?1:0;
                if(!node->exists(x)){
                    node->set(x,new Node());
                }
                node = node->next(x);
            }
        }

        int maxi = 0;
        for(int i=0;i<nums.size();i++){
            int num = 0;
            int a = nums[i];
            Node* node = root;
            for(int j=31;j>=0;j--){
                int x = a&(1<<j)?1:0;
                if(node->exists(!x)){
                    num+= (1<<j);
                    x=!x;
                }
                node= node->next(x);
            }
            maxi = max(maxi,num);
        }


        return maxi;
    }
};
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