<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:
        
        
        t = {}
        for n in nums:
            p = t
            for i in range(31,-1,-1):
                mask = 1 << i
                d = 1 if mask & n else 0
                p[d] = p.get(d, {})
                p = p[d]
            p['$'] = n
            
        res = 0
        
        for n in nums:
            p = t
            for i in range(31,-1,-1):
                mask = 1 << i
                d = 1 if mask & n else 0
                
                if d == 0:
                    if 1 in p:
                        p = p[1]
                    else:
                        p = p[0]
                else:
                    if 0 in p:
                        p = p[0]
                    else:
                        p = p[1]
                        
            res = max(res, p['$'] ^ n)
            
        return res
                        
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