<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:
        l=0
        r=len(nums)-1
        val=-1
        z=(l+r)//2
        def checking(l,r):
            if l<=r:
                mid=(l+r)//2
                if mid==len(nums)-1:
                    if nums[mid-1]!=nums[mid]:
                        return mid
                elif mid==0:
                    if nums[mid]!=nums[mid+1]:
                        return mid
                elif nums[mid]!=nums[mid+1] and nums[mid]!=nums[mid-1]:
                    return mid
                p=checking(mid+1,r)+checking(l,mid-1)
                return p
            else:
                return 0
        l=checking(l,r)
        return nums[l]
26900