<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:
        if len(nums) < 2:
            return nums[0]

        l = 0
        r = len(nums) - 1

        while l < r:
            
            m = l + (r - l) // 2
            print(l, m, r)
            if nums[m - 1] != nums[m] != nums[m + 1]:
                return nums[m]

            if m + 1 < len(nums) and nums[m] == nums[m + 1]:
                if m % 2:
                    r = m - 1
                else:
                    l = m + 1
            else:
                if m % 2:
                    l = m + 1
                else:
                    r = m - 1
            
        return nums[l] 
26100