<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 TrieNode:
    def __init__(self):
        self.children = [None, None]

class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, n):
        cur = self.root
        for i in range(31, -1, -1):
            cur_bit = (n >> i) & 1
            if not cur.children[cur_bit]:
                cur.children[cur_bit] = TrieNode()
            cur = cur.children[cur_bit]
    
    def get_max_xor(self, n):
        cur = self.root
        other_num = 0
        for i in range(31, -1, -1):
            req_bit = 1 - ((n >> i) & 1)
            if cur.children[req_bit]:
                other_num = (other_num << 1) | req_bit
                cur = cur.children[req_bit]
            else:
                other_num = (other_num << 1) | (1 - req_bit)
                cur = cur.children[1 - req_bit]
        return other_num ^ n

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        trie = Trie()
        for n in nums:
            trie.insert(n)
        
        res = 0
        for n in nums:
            res = max(res, trie.get_max_xor(n))
        
        return res
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