<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.chd = {}

    def add(self, num):
        node = self
        for digit in num:
            if digit not in node.chd:
                node.chd[digit] = TrieNode()
            node = node.chd[digit]

    def remove(self, num):
        node = self
        stack = []
        for digit in num:
            stack.append([node, digit])
            node = node.chd[digit]

        node, digit = stack.pop()
        node.chd.pop(digit)
        if len(node.chd) == 0:
            while stack:
                node, digit = stack.pop()
                node.chd.pop(digit)
                if len(node.chd):
                    break


class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        nums = list(set(nums))
        nums.sort()
        n = len(nums)
        L = max(1, max(nums).bit_length())

        def inv(x):
            if x == "1":
                return "0"
            return "1"

        root = TrieNode()
        ans = 0
        l = 0
        for r in range(n):
            cur = bin(nums[r])[2:]
            cur = f"{cur:0>{L}}"

            while nums[l] < nums[r] / 2:
                prv = bin(nums[l])[2:]
                prv = f"{prv:0>{L}}"
                root.remove(prv)
                l += 1

            node = root
            res = 0
            for j in range(L):
                res <<= 1
                if inv(cur[j]) in node.chd:
                    res += 1
                    node = node.chd[inv(cur[j])]
                elif cur[j] in node.chd:
                    node = node.chd[cur[j]]

            ans = max(ans, res)

            root.add(cur)

        return ans
80700