<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:
    maxNum = max(nums)
    if maxNum == 0:
      return 0
    maxBit = int(math.log2(maxNum))
    ans = 0
    mask = 0

    for i in range(maxBit, -1, -1):
      mask |= 1 << i
      prefixes = set([num & mask for num in nums])
      candidate = ans | 1 << i
      for prefix in prefixes:
        if prefix ^ candidate in prefixes:
          ans = candidate
          break

    return ans
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