<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]
        self.num = 0

class Trie:
    def __init__(self, nums, maxlen):
        self.root = TrieNode()
        for n in nums:
            curr = self.root
            for i in range(maxlen, -1, -1):
                val = 1 if (1 << i) & n else 0
                if not curr.children[val]: curr.children[val] = TrieNode()
                curr = curr.children[val]
            curr.num = n

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        maxval = len(bin(max(nums))) - 2
        trie = Trie(nums, maxval)
        ans = 0
        for n in nums:
            curr = trie.root
            for i in range(maxval, -1, -1):
                val = 1 if (1 << i) & n else 0
                curr = curr.children[1 - val] if curr.children[1 - val] else curr.children[val]
            ans = max(ans, n ^ curr.num)
        return ans
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