<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.v = [0]
        self.lt = [-1]
        self.rt = [-1]
        self.id = 0

    def create(self):
        self.v.append(0)
        self.lt.append(-1)
        self.rt.append(-1)
        self.id += 1

    def add(self, x):
        node = 0
        for bit in range(30, -1, -1):
            b = x & (1 << bit)
            self.v[node] += 1

            if b == 0:
                if self.lt[node] == -1:
                    self.create()
                    self.lt[node] = self.id
                node = self.lt[node]
            else:
                if self.rt[node] == -1:
                    self.create()
                    self.rt[node] = self.id
                node = self.rt[node]
        self.v[node] += 1

    def maxxor(self, x):
        node = 0
        ret = 0
        for bit in range(30, -1, -1):
            b = (x >> bit) & 1

            if b == 0:
                if self.rt[node] != -1 and self.v[self.rt[node]] > 0:
                    ret += 1 << bit
                    node = self.rt[node]
                else:
                    node = self.lt[node]
            else:
                if self.lt[node] != -1 and self.v[self.lt[node]] > 0:
                    ret += 1 << bit
                    node = self.lt[node]
                else:
                    node = self.rt[node]
        return ret

    def erase(self, x):
        node = 0
        for bit in range(30, -1, -1):
            b = x & (1 << bit)
            self.v[node] -= 1

            if b == 0:
                node = self.lt[node]
            else:
                node = self.rt[node]
        self.v[node] -= 1


class Solution:
    def maximumStrongPairXor(self, nums):
        T = Trie()
        nums.sort(reverse=True)
        ans = 0
        ptr = 0
        for x in nums:
            T.add(x)
            while ptr < len(nums) and nums[ptr] > 2 * x:
                T.erase(nums[ptr])
                ptr += 1
            ans = max(ans, T.maxxor(x))
        return ans
40700