<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:
//     long singleNonDuplicate(vector<int>& nums) {
       
//         long 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];
//         long l =0,r =n-1;
//         long ans =0;
//         while(l<=r){
//             long mid = l+(r-l)/2;
//             if(mid > 0 && mid < n - 1 && nums[mid]!=nums[mid-1] && nums[mid]!=nums[mid+1]){
//                ans =nums[mid];
//                break;
//             }
//             if(((mid-l+1)%2!=0 && nums[mid]==nums[mid-1]) || ((mid-l+1)%2==0 && nums[mid]==nums[mid+1])){
//                 r =mid-1;
//             }else{
//                 l =mid+1;
//             }
//         }
        
//         return ans;    
//     }
// };

class Solution {
public:
    long singleNonDuplicate(vector<int>& nums) {
        ios_base::sync_with_stdio(false);  // Correct usage
        cin.tie(nullptr);  // Proper stream tie
        cout.tie(nullptr);

        long 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];

        long l = 0, r = n - 1;
        long ans = 0;
        
        while (l <= r) {
            long mid = l + (r - l) / 2;

            // Ensure mid is within bounds for comparison
            if (mid > 0 && mid < n - 1 && nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1]) {
                ans = nums[mid];
                break;
            }

            if ((mid % 2 == 0 && nums[mid] == nums[mid + 1]) || (mid % 2 == 1 && nums[mid] == nums[mid - 1])) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }

        return ans;    
    }
};

25200