<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* links[2];
};
class Trie{
    public:
    Node* root;
    Trie(){
        root = new Node();
    }
    void insert(string word){
        int n = word.length();
        Node * curr = root;
        for(int i=0;i<n;i++){
            if(curr->links[word[i]-'0']==NULL){
                curr->links[word[i]-'0'] = new Node();
            }
            curr = curr->links[word[i]-'0'];
        }
    }
};
class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        int n = nums.size();
        Trie t;
        for(int i=0;i<n;i++){
            int k = nums[i];
            string s;
            while(k){
                s+=(to_string(k%2));
                k= k>>1;
            }
            while(s.length()<32){
                s+='0';
            }
            reverse(s.begin(),s.end());
            t.insert(s);
        }
        int mx =0;
        for(int i=0;i<n;i++){
            int k = nums[i];
            string s;
            while(k){
                s+=(to_string(k%2));
                k/=2;
            }
            while(s.length()<32){
                s+='0';
            }
            reverse(s.begin(),s.end());
            int val=0;
            Node *curr = t.root;
            //cout<<s<<endl;
            for(int j=0;j<32;j++){
                int bit = s[j] - '0';
                if (curr->links[1 - bit]) {
                    val = (val << 1) | 1;
                    curr = curr->links[1 - bit];
                } else {
                    val = val << 1;
                    curr = curr->links[bit];
                }
            }
            mx = max(mx,val);
        }
        return mx;
    }
};
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