<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:
        L = len(bin(max(nums))) - 2
        for i in range(len(nums)):
            b = bin(nums[i])[2:]
            nums[i] = '0' * (L - len(b)) + b

        max_xor = 0
        trie = {}

        for num in nums:
            node = trie
            xor_node = trie
            curr_xor = 0
            for bit in num:
                if bit not in node:
                    node[bit] = {}
                node = node[bit]

                toggled = '1' if bit is '0' else '0'
                if toggled in xor_node:
                    curr_xor = (curr_xor << 1) | 1
                    xor_node = xor_node[toggled]
                else:
                    curr_xor <<= 1
                    xor_node = xor_node[bit]
            max_xor = max(curr_xor, max_xor)
        
        return max_xor
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