<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 Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        def bit(n, i):
            return (n & (1 << i)) >> i
        n = len(nums)
        nums.sort()
        c = 0
        for i in range(n):
            #find leftmost element > 2 * nums[i]
            l, r = i, n
            while l < r - 1:
                m = (l + r) // 2
                if nums[m] <= 2 * nums[i]:
                    l = m
                else:
                    r = m

            #invariant: if we xor the bits from 20 to j of nums[i] with anything in nums[gl : gr],
            #they will all give the same current maximal xor with nums[i]
            gl, gr = i, r
            for j in range(20, -1, -1):
                if gl == gr - 1:
                    #gl is our solution
                    break
                bj = bit(nums[i], j)
                bgl = bit(nums[gl], j)
                if bgl == bit(nums[gr - 1], j):
                    #bit j for this entire range is the same
                    #hence, we won't be able to reduce the size of the range this iteration
                    continue

                l, r = gl, gr
                while l < r - 1:
                    m = (l + r) // 2
                    if bit(nums[m], j) == bgl:
                        l = m
                    else:
                        r = m
                #find the position where the jth bit transitions from 1 to 0 or 0 to 1

                if bj != bgl:
                    #then we care about left range
                    gl, gr = gl, r
                else:
                    #else right range
                    gl, gr = r, gr

            c = max(c, nums[i] ^ nums[gl])
        return c
24000