<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, val=-1):
        self.val = val
        self.left = None
        self.right = None

class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, num):
        curr = self.root
        for i in range(31, -1, -1):
            ith_bit = num & (1 << i)
            if ith_bit == 0:
                if curr.left == None:
                    curr.left = TrieNode(0)
                curr = curr.left
            else:
                if curr.right == None:
                    curr.right = TrieNode(1)
                curr = curr.right
    
    def findMaxXor(self, num):
        xor = 0
        curr = self.root
        for i in range(31, -1, -1):
            ith_bit = num & (1 << i)
            if ith_bit == 0:
                if curr.right != None:
                    xor += (1 << i)
                    curr = curr.right
                else:
                    curr = curr.left
            else:
                if curr.left != None:
                    xor += (1 << i)
                    curr = curr.left
                else:
                    curr = curr.right
        return xor
     
class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        maxXorFinder = Trie()
        ans = 0
        for i in nums:
            maxXorFinder.insert(i)
            ans = max(ans, maxXorFinder.findMaxXor(i))
        return ans
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