<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 PrefixTree:
    def __init__(self):
        self.children = {}
    
    def add(self, word):
        p = self
        for c in word:
            if c not in p.children:
                p.children[c] = PrefixTree()
            p = p.children[c]

class Solution:
    # nums: 31 bit ints
    def findMaximumXOR(self, nums: List[int]) -> int:
        prefix_tree = PrefixTree()
        res = 0
        for x in nums:
            x_bin = bin(x)[2:].rjust(31, "0")
            prefix_tree.add(x_bin)
            xor = 0
            p = prefix_tree
            for i, b in enumerate(x_bin):
                neg_b = "1" if b == "0" else "0"
                if neg_b in p.children:
                    p = p.children[neg_b]
                    xor |= (1 << (31 - i - 1))
                else:
                    p = p.children[b]
            res = max(res, xor)
        return res
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