<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 TrieNode:
    def __init__(self):
        self.children = [None]*2
        self.is_end = False
        self.num = None


class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word: str) -> None:
        curr = self.root
        for i in range(64,-1,-1):
            indx = 1 if(1<<i) & word else 0
            if not curr.children[indx]:
                curr.children[indx] = TrieNode()
            curr = curr.children[indx]
        curr.is_end = True
        curr.num = word
    def find(self,num):
        curr = self.root
        for i in range(64,-1,-1):
            indx = 0 if(1<<i)&num else 1
            if curr.children[indx]:
                curr = curr.children[indx]
            else:
                curr = curr.children[indx^1]
        return num^curr.num
                

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        trie = Trie()
        ans = 0
        for num in nums:
            trie.insert(num)
        for num in nums:
            ans = max(ans,trie.find(num))
        return ans
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