<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 Tri():
    def __init__(self):
        self.arr=[None,None]
    
    def insert(self,num):
        tmp=self.arr
        a=[]
        tnum=num
        for i in range(32):
            a.append(tnum%2)
            tnum//=2
        a=a[::-1]
        for i in a:
            if tmp[i]==None:
                tmp[i]=Tri().arr
            tmp=tmp[i]    
    def findmx(self,num):
        tmp=self.arr
        a=[]
        tnum=num
        for i in range(32):
            a.append(tnum%2)
            tnum//=2
        a=a[::-1]
        ans=[]
        for i in a:
            if tmp[(i+1)%2]!=None:
                ans.append(1)
                tmp=tmp[(i+1)%2]
            else:
                ans.append(0)
                tmp=tmp[i]
        anss=0
        cnt=0
        for i in ans[::-1]:
            anss+=(2**cnt)*i
            cnt+=1
        return anss


class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:

        tri=Tri()
        for i in nums:
            tri.insert(i)
        ans=0
        for i in nums:
            ans=max(ans,tri.findmx(i))
        return ans


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