<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>

from typing import List

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        max_xor = 0
        mask = 0
        
        # Iterate over each bit position from 31 to 0
        for i in range(31, -1, -1):
            # Update the mask to include the current bit
            mask |= (1 << i)
            found_prefixes = set()
            
            # Collect all prefixes of length (31 - i)
            for num in nums:
                found_prefixes.add(num & mask)
            
            # Start with a potential new max_xor by setting the current bit
            potential_max_xor = max_xor | (1 << i)
            
            # Check if there exists two prefixes that can achieve this potential_max_xor
            for prefix in found_prefixes:
                if (prefix ^ potential_max_xor) in found_prefixes:
                    max_xor = potential_max_xor
                    break
        
        return max_xor

# Example usage:
# sol = Solution()
# print(sol.findMaximumXOR([3,10,5,25,2,8]))  # Output: 28
# print(sol.findMaximumXOR([14,70,53,83,49,91,36,80,92,51,66,70]))  # Output: 127
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