<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>

def split(nums, k):
    a, b = [], []
    for val in nums:
        if val & (1 << k):
            a.append(val)
        else:
            b.append(val)
    return a, b

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        m = max(nums).bit_length()
        trie = {}
        for val in nums:
            cur = trie
            for i in reversed(range(m)):
                if val & (1 << i):
                    cur[1] = cur.get(1, {})
                    cur = cur[1]
                else:
                    cur[0] = cur.get(0, {})
                    cur = cur[0]
            cur[0] = val
        ans = 0
        for val in nums:
            cur = trie
            for i in reversed(range(m)):
                if 1 not in cur:
                    cur = cur[0]
                elif 0 not in cur:
                    cur = cur[1]
                elif val & (1 << i):
                    cur = cur[0]
                else:
                    cur = cur[1]
            ans = max(ans, val^cur[0])
        return ans

        
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