Maximum XOR of Two Numbers in an Array
<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>
Bit Manipulation; Trie; Array; Hash Table
class Trie:
    __slots__ = ("children",)

    def __init__(self):
        self.children: List[Trie | None] = [None, None]

    def insert(self, x: int):
        node = self
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v] is None:
                node.children[v] = Trie()
            node = node.children[v]

    def search(self, x: int) -> int:
        node = self
        ans = 0
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v ^ 1]:
                ans |= 1 << i
                node = node.children[v ^ 1]
            else:
                node = node.children[v]
        return ans


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