<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];
    bool flag = 0;
    bool is_contain(int ind){
        return links[ind]!=NULL;
    }
};
class Trie{
    Node* root;
public:
    Trie(){
        root = new Node();
    }
    void insert(int val){
        Node* tmp = root;
        for(int i=31;i>=0;i--){
            int mask = ((val>>i)&1);
            if(!tmp->is_contain(mask)){
                tmp->links[mask] = new Node();
            }
            tmp = tmp->links[mask];
        }
        tmp->flag = 1;
    }
    int max_xor(int val){
        int ans = 0;
        Node* tmp = root;
        for(int i=31;i>=0;i--){
            int mask = ((val>>i)&1);
            if(tmp->is_contain(mask^1)){
                ans+=(1<<i);
                tmp = tmp->links[mask^1];
            }
            else{
                tmp = tmp->links[mask];
            }
            if(tmp==NULL) break;
        }
        return ans;
    }
};
class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        Trie trie;
        for(auto &it:nums){
            trie.insert(it);
        }
        int ans = 0;
        for(auto &it:nums){
            ans = max(ans,trie.max_xor(it));
        }
        return ans;
    }
};
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