<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,None]
class Trie :
    def __init__(self):
        self.root = TrieNode()
    def insert(self,val):
        curr = self.root
        for i in reversed(range(32)):
            if (1<<i)&val:
                if curr.children[1]==None:
                    curr.children[1] = TrieNode()
                curr = curr.children[1]
            else:
                if curr.children[0]==None:
                    curr.children[0]=TrieNode()        
                curr = curr.children[0]
    def find_max(self,val):
        curr = self.root
        ans = 0
        for i in reversed(range(32)):
            if (1<<i)&val:
                if  curr.children[0]!=None:
                    curr = curr.children[0]
                    ans+= 2**i
                else:
                    curr = curr.children[1]
            else:
                if curr.children[1]!=None:
                    curr = curr.children[1]
                    ans+=2**i

                else:
                    curr = curr.children[0]
        return ans

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        t = Trie()
        maxi = 0
        t.insert(nums[0])
        for x in nums[1:]:
            maxi  = max(maxi , t.find_max(x))
            t.insert(x)
        return maxi
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