<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>

from dataclasses import dataclass, field

factory = lambda : [None, None]

@dataclass
class Node:
    children: List[Optional['Node']] = field(default_factory=factory)

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

    def insert(self, number: int):
        cur = self.root
        for i in range(31, -1 , -1):
            e_int = (number >> i) & 1
            if not cur.children[e_int]:
                cur.children[e_int] = Node()
            cur = cur.children[e_int]

    def max_xor(self, number):
        cur = self.root
        max_xor = 0
        for i in range(31, -1, -1):
            e_int = (number >> i) & 1
            next_bit = 1 ^ e_int  # alternate bit i.e 0 if e_int was 1 and vice-versa
            if cur.children[next_bit]:
                max_xor |= (1 << i)
                cur = cur.children[next_bit]
            else:
                #  no need to update ans as no alternating next bit found, wouldnt affect the result
                cur = cur.children[e_int]
        return max_xor

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

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