<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 Trie:
    def __init__(self, l):
        self.data = {}
        self.l = l
    def insert(self, x):
        node = self.data
        xor_node = self.data
        xor = 0
        for bit in x:
            if bit not in node:
                node[bit] = {}
            node = node[bit]

            toggle_bit = 1 - bit
            if toggle_bit in xor_node:
                xor = xor << 1 | 1
                xor_node = xor_node[toggle_bit]
            else:
                xor = xor << 1
                xor_node = xor_node[bit]
        return xor



class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        L = len(bin(max(nums)))-2

        bin_nums = [[(x >> i) & 1 for i in range(L)][::-1] for x in nums]

        #print(bin_nums)
        trie = Trie(L)
        res = 0
        for num in bin_nums:
            xor = trie.insert(num)
            res = max(res, xor)
        #res = 0
        #for num in bin_nums:
        #    res = max(res, trie.max_xor(num))

        return res
646528