<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 = {}

class BinaryNumDictionary:
    def __init__(self):
        self.root = TrieNode()
    
    def insert_num(self, num):
        bin_num_str = bin(num)[2:].zfill(32)
        curr = self.root
        for c in bin_num_str:
            if c not in curr.children:
                curr.children[c] = TrieNode()
            curr = curr.children[c]

    def find_xor(self, num):
        bin_num_str = bin(num)[2:].zfill(32)

        xor_str = ""

        curr = self.root
        for c in bin_num_str:
            if c == "0":
                if "1" in curr.children:
                    xor_str += "1"
                    curr = curr.children["1"]
                else:
                    xor_str += "0"
                    curr = curr.children["0"]
            else:
                if "0" in curr.children:
                    xor_str += "0"
                    curr = curr.children["0"]
                else:
                    xor_str += "1"
                    curr = curr.children["1"]

        return int(xor_str, 2) ^ num



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

        d = BinaryNumDictionary()
        for num in nums:
            d.insert_num(num)
        
        max_xor = 0

        for num in nums:
            max_xor = max(max_xor, d.find_xor(num))

        return max_xor

        # maxXOR = 0
        # mask = 0
        
        # for i in range(31, -1, -1):  # Iterate from MSB (31st bit) to LSB (0th bit)
        #     mask |= (1 << i)  # Update the mask to include the current bit
        #     prefixes = set()
            
        #     for num in nums:
        #         # Keep only the leftmost bits up to the current bit
        #         prefixes.add(num & mask)
            
        #     # Assume the current bit is 1 in the maximum XOR result
        #     newMaxXOR = maxXOR | (1 << i)
            
        #     # Check if there are two prefixes that can achieve this newMaxXOR
        #     for prefix in prefixes:
        #         if (prefix ^ newMaxXOR) in prefixes:
        #             maxXOR = newMaxXOR
        #             break
        
        # return maxXOR        
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