<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 TrieNode:
    def __init__(self):
        self.childs = [None] * 2
        self.cnt = 0

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

    def add(self, num):
        p = self.root
        for i in range(31, -1, -1):
            cur = 1 if (num & (1 << i)) else 0
            if not p.childs[cur]:
                p.childs[cur] = TrieNode()
            p = p.childs[cur]
            p.cnt += 1

    def remove(self, num):
        p = self.root
        for i in range(31, -1, -1):
            cur = 1 if (num & (1 << i)) else 0
            p.childs[cur].cnt -= 1
            if p.childs[cur].cnt == 0:
                p.childs[cur] = None
                break
            p = p.childs[cur]

    def query(self, target):
        p = self.root
        res = 0
        for i in range(31, -1, -1):
            cur = 1 if (target & (1 << i)) else 0
            xor_cur = cur ^ 1
            if p.childs[xor_cur]:
                p = p.childs[xor_cur]
                res |= (1 << i)
            else:
                p = p.childs[cur]

        return res
class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        N = len(nums)
        nums.sort()
        tree = Trie()
        left = 0
        res = 0
        for i in range(N):
            while left < i:
                if nums[i] - nums[left] > nums[left]:
                    tree.remove(nums[left])
                    left += 1
                else:
                    break
            if left < i:
                res = max(res, tree.query(nums[i]))

            tree.add(nums[i])
        return res

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