<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 {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int n = nums.size();
        if(n==1) return nums[0];
        if(nums[0]!=nums[1]) return nums[0];
        if(nums[n-1]!= nums[n-2]) return nums[n-1];
        int low =1 ; int high = n-2;
        while(low<=high){
            int mid =(low+high)/2;
            if(nums[mid]!=nums[mid+1]&& nums[mid]!= nums[mid-1]){
                return nums[mid];
            }

            //we are in left
            if((mid % 2== 1 && nums[mid]== nums[mid-1])
            || (mid % 2 == 0 && nums[mid] == nums[mid+1])){
                low = mid+1;
            }
            //we are in right
            else {
                high = mid -1;
            }
        }
        return -1;
        
        
    }
};
24800