<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:
        # Length of the max number in a binary representation
        L = len(bin(max(nums))) - 2
        new = [bin(num)[2:] for num in nums]
        max_xor = 0
        for i in range(L)[::-1]:
            # Go to the next bit by the left shift
            max_xor <<= 1

            # Set 1 in the smallest bit
            curr_xor = max_xor | 1

            # Compute all existing prefixes 
            # of length (L - i) in binary representation
            prefixes = {num >> i for num in nums}

            # Update max_xor, if two of these prefixes could result in curr_xor.
            # Check if p1^p2 == curr_xor, i.e. p1 == curr_xor^p2
            max_xor |= any(curr_xor^p in prefixes for p in prefixes)
                    
        return max_xor
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