<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]  # 0 and 1

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, num):
        node = self.root
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            if not node.children[bit]:
                node.children[bit] = TrieNode()
            node = node.children[bit]

    def findMaxXOR(self, num):
        node = self.root
        max_xor = 0
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            # To maximize XOR, we want to go to the opposite bit if it exists
            if node.children[1 - bit]:
                max_xor |= (1 << i)
                node = node.children[1 - bit]
            else:
                node = node.children[bit]
        return max_xor

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

# Example usage:
solution = Solution()
nums = [3, 10, 5, 25, 2, 8]
print(solution.findMaximumXOR(nums))  # Output: 28
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