<p>You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.</p>

<p>Return <em>the single element that appears only once</em>.</p>

<p>Your solution must run in <code>O(log n)</code> time and <code>O(1)</code> space.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,2,3,3,4,4,8,8]
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [3,3,7,7,10,11,11]
<strong>Output:</strong> 10
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        lo = 0
        hi = len(nums) - 1

        while lo <= hi:
            md = lo + (hi - lo) // 2

            if (
                md % 2 == 0
                and md + 1 < len(nums)
                and nums[md] == nums[md + 1]
                or md % 2 != 0
                and md - 1 >= 0
                and nums[md] == nums[md - 1]
            ):
                lo = md + 1
            else:
                hi = md - 1
        return nums[lo]

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