<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 Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        self.trie = {}

        # conver numbers to binary repr w/ same length
        bins = [bin(x)[2:] for x in nums]
        max_len = max([len(x) for x in bins])
        bins = [f'{x.rjust(max_len, "0")}' for x in bins]

        # construct trie
        for b in bins:
            cur = self.trie
            for char in b:
                if char not in cur:
                    cur[char] = {}
                cur = cur[char]
            cur['val'] = b
        
        diff = {'0': '1', '1': '0'}
        max_xor = 0
        for b in bins:
            cur = self.trie
            for char in b:
                cur = cur[char]
            num1 = cur['val']
    
            cur = self.trie
            for char in num1:
                if diff[char] in cur:
                    cur = cur[diff[char]]
                else:
                    cur = cur[char]
            num2 = cur['val']
    
            max_xor = max(max_xor, int(num1, 2) ^ int(num2, 2))
        return max_xor
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