<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* children[2];
    int number;
    bool is_num = false;
    void insert(int x) {
        auto cur = this;
        bitset<32> str(x);

        for(int i = 31; i >= 0; i--) {
            if ( !cur->children[!str[i]] ) cur->children[!str[i]] = new Trie();
            cur = cur->children[!str[i]];
        }
        cur->number = x;
        cur->is_num = true; 
    }

    int find(int x) {
        auto cur = this;
        bitset<32> str(x);
        for(int i = 31; i >= 0; i--) {
            int digit = !str[i];
            if (!cur->children[digit]) return -1;
            cur = cur->children[digit];
        }

        return cur->number;
    }

    int get_max_xor(int x) {
        auto cur = this;
        bitset<32> str(x);
        for(int i = 31; i >= 0; i--) {
            if(!str[i]) {
               cur = cur->children[0] ? cur->children[0] : cur->children[1];
            } else {
                cur = cur->children[1] ? cur->children[1] : cur->children[0];
            }
        }
        return cur->number ^ x;
    }
    
};
class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        int max_xor = 0;
        auto trie = new Trie();
        for (int i = 0; i < nums.size(); i++) {
           trie->insert(nums[i]);
           max_xor = max(max_xor, trie->get_max_xor(nums[i]));
        }
        return max_xor;
    }
};
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