<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 Node:
    def __init__(self):
        self.arr=[None]*2

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

    def insert(self,a):
        curr=self.root
        for i in range(31,0,-1):
            x=1<<(i-1)
            if x&a==0:
                ind=0
            else:
                ind=1
            if curr.arr[ind] is None:
                curr.arr[ind]=Node()
            curr = curr.arr[ind]           

    def getMaxXor(self,a):
        curr=self.root
        ans=0
        for i in range(31,0,-1):
            x=1<<(i-1)
            if x&a==0:
                get=1
            else:
                get=0
            if curr.arr[get] is None:
                curr=curr.arr[1-get]
            else:
                curr=curr.arr[get]
                ans+=1<<(i-1)
        return ans


class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        t=Trie()
        res=0
        for a in nums:
            t.insert(a)

        for a in nums:
            res=max( res , t.getMaxXor(a)  )
        return res

        
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