<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:
        return self.trie(nums)
    
    def trie(self, A):
        if not A:
            return -1
        
        self.length = 0
        self.result = 0
        self.tree = {}
        binA = [bin(num)[2:] for num in A]
        for num in binA:
            self.length = max(len(num), self.length)
        
        for num in binA:
            self.build(self.length, num, self.tree)

        for num in binA:
            self.findxor(num, self.tree, "")
        
        return self.result
    
    def build(self, length, num, node):        
        if not num:
            return
        
        if length > len(num):
            if 0 not in node:
                node[0] = {}
            node = node[0]
            self.build(length-1, num, node)
            return
        
        bit = int(num[0])
        if bit not in node:
            node[bit] = {}
        node = node[bit]
        self.build(length-1, num[1:], node)
    
    def findxor(self, num, node, cur):
        if not num:
            self.result = max(self.result, int(cur, 2))
            return
        
        bit = int(num[0])
        if 1-bit in node:
            node = node[1-bit]
            self.findxor(num[1:], node, cur + "1")
        else:
            node = node[bit]
            self.findxor(num[1:], node, cur + "0")
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