<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 Trie{
    public:
        Trie* child[2];
        bool is_leaf;

        Trie(){
            this->is_leaf=false;
            this->child[0]=NULL;
            this->child[1]=NULL;
        }
};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        int n=nums.size();

        Trie* root=new Trie();
        for(int i=0;i<n;i++){
            Trie* node=root;
            for(int j=30;j>=0;j--){
                int mask=(1<<j);
                int bit=((mask&nums[i])==mask);
                if(node->child[bit]==NULL){
                    node->child[bit]=new Trie();
                }

                node=node->child[bit];
            }
        }

        int max_X=0;
        for(int i=0;i<n;i++){
            int maxi=0;
            Trie* node=root;
            for(int j=30;j>=0;j--){
                int mask=(1<<j);
                int bit=!((mask&nums[i])==mask);
                if(node->child[bit]!=NULL){
                    maxi+=mask;
                    node=node->child[bit];
                }
                else{
                    node=node->child[!bit];
                }
            }

            max_X=max(max_X,maxi);
        }

        return max_X;
    }
};
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