<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, arr: List[int]) -> int:
        pass
        n=len(arr)
        if(n==1): return arr[0]
        if arr[0]!=arr[1]: return arr[0]
        if (arr[n-1]!=arr[n-2]):
            return arr[n-1]
        low=1
        high=n-2

        while(low<=high):
            mid=(low+high)//2
            if arr[mid]!=arr[mid-1] and arr[mid]!=arr[mid+1]:
                return arr[mid]
            if (mid%2==1 and arr[mid]==arr[mid-1]) or (mid%2==0 and arr[mid]==arr[mid+1]):
                low=mid+1
            else:
                high=mid-1
        return -1
26700