<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. A pair of integers <code>x</code> and <code>y</code> is called a <strong>strong</strong> pair if it satisfies the condition:</p>

<ul>
	<li><code>|x - y| &lt;= min(x, y)</code></li>
</ul>

<p>You need to select two integers from <code>nums</code> such that they form a strong pair and their bitwise <code>XOR</code> is the <strong>maximum</strong> among all strong pairs in the array.</p>

<p>Return <em>the <strong>maximum</strong> </em><code>XOR</code><em> value out of all possible strong pairs in the array</em> <code>nums</code>.</p>

<p><strong>Note</strong> that you can pick the same integer twice to form a pair.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 11 strong pairs in the array <code>nums</code>: (1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) and (5, 5).
The maximum XOR possible from these pairs is 3 XOR 4 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,100]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are 2 strong pairs in the array nums: (10, 10) and (100, 100).
The maximum XOR possible from these pairs is 10 XOR 10 = 0 since the pair (100, 100) also gives 100 XOR 100 = 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [500,520,2500,3000]
<strong>Output:</strong> 1020
<strong>Explanation:</strong> There are 6 strong pairs in the array nums: (500, 500), (500, 520), (520, 520), (2500, 2500), (2500, 3000) and (3000, 3000).
The maximum XOR possible from these pairs is 500 XOR 520 = 1020 since the only other non-zero XOR value is 2500 XOR 3000 = 636.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2<sup>20</sup> - 1</code></li>
</ul>

class Trie:

    def __init__(self, max_bits):
        self.children = [None, None]
        self.max_bits = max_bits
        self.end = 0
    
    def add(self, num):
        node = self
        for b in reversed(range(self.max_bits)):
            bit = (num >> b) & 1
            if not node.children[bit]:
                node.children[bit] = Trie(self.max_bits)
            node = node.children[bit]
        node.end += 1
    
    def delete(self, num):
        node = self
        stack = []
        for b in reversed(range(self.max_bits)):
            bit = (num >> b) & 1
            child = node.children[bit]
            stack.append([node, child, bit])
            node = child
        node.end -= 1
        while stack and stack[-1][1].end == 0 and all(not child for child in stack[-1][1].children):
            parent, child, bit = stack.pop()
            parent.children[bit] = None

    def find_max(self, y):
        XOR = 0
        node = self
        for b in reversed(range(self.max_bits)):
            bit = (y >> b) & 1
            if node.children[1 - bit]:
                XOR |= 1 << b
                node = node.children[1 - bit]
            else:
                node = node.children[bit]
        
        return XOR

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        max_XOR = 0
        max_bits = max(int(log(n, 2)) + 1 for n in nums)
        trie = Trie(max_bits)
        nums.sort()
        Q = deque()

        for n in nums:
            trie.add(n)
            Q.append(n)
            while Q and 2 * Q[0] < n:
                trie.delete(Q.popleft())
            max_XOR = max(max_XOR, trie.find_max(n))
        
        return max_XOR
63200