<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):
        self.root = {}

    def add(self, v):
        p = self.root
        for i in range(19, -1, -1):
            f = (v >> i) & 1
            if f not in p:
                p[f] = [{}, 1]
            else:
                p[f][1] += 1
            p = p[f][0]

    def remove(self, v):
        p = self.root
        for i in range(19, -1, -1):
            f = (v >> i) & 1
            p[f][1] -= 1
            if p[f][1] == 0:
                del p[f]
                break
            p = p[f][0]
            
    def find(self, v):
        p = self.root
        r = 0
        for i in range(19, -1, -1):
            f = (v >> i) & 1
            g = 1 - f
            if g in p:
                r |= g << i
                p = p[g][0]
            elif f in p:
                r |= f << i
                p = p[f][0]
            else:
                return r
        return r

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        ans = 0
        trie = Trie()
        q = deque()
        for v in sorted(nums):
            while q and q[0] * 2 < v:
                trie.remove(q.popleft())
            trie.add(v)
            q.append(v)
            ans = max(ans, trie.find(v) ^ v)
        return ans
81100