<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 bittrie():
    def __init__(o, n):
        o.k = len(bin(n))-2
        o.bcnt = [0]*(2**(o.k+1))
    def _addrem(o, v, d):
        ci = 1
        for i in range(o.k-1,-1,-1):
            if v & (1<<i): ci = ci*2+1
            else: ci *= 2
            o.bcnt[ci] += d
    def rem(o, v): o._addrem(v,-1)
    def add(o, v): o._addrem(v,1)
    def findmax(o,v):
        cmax, i, ci = 0, o.k-1, 1
        if o.bcnt[2] == 0 and o.bcnt[3] == 0: return 0
        for i in range(o.k-1, -1, -1):
            if v & (1<<i): ci, oi = ci*2+1, 2*ci
            else: ci, oi = 2*ci, 2*ci+1
            if o.bcnt[oi] > 0: cmax, ci = cmax | (1<<i), oi
            #print(bin(v)[2:], 'i', i, 'cmax ', bin(cmax)[2:])
        return cmax

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        nums.sort()
        res, r, n, bt = 0, 0, len(nums), bittrie(nums[-1])
        for i in range(n-1):
            v = nums[i]
            while r < n and nums[r] <= 2*v: 
                bt.add(nums[r])
                r += 1
            #print((i,v), 'r ', r, bt.bcnt[:16])
            res = max(res, bt.findmax(nums[i]))
            bt.rem(nums[i])
        return res
40400