Binary search in sorted array, Medium - Problem Description - You are given a sorted array consisting of only integers - where every element appears exactly twice, - except for one element which appears exactly once. - Find this single element that appears only once. - Follow up - Your solution should run in O(logN) time and O(1) space. - Examples

<summary>Example </summary>

    Input: nums = [3,3,7,7,10,11,11] <br>
    Output: 10
  </details>
- Approach
  - It is evident from the input above that a new number starts from even index and ends at immediate next index i.e. next odd index for example,if the given `array` is `[1,1,2,2,3,3,4,4]` new element can be seen at `index: 0,2,4,6` (Note:Array consists of duplets only)
  - But **WHAT IF** array consists of one single element as well? example:`[1,1,2,2,3,4,4]` Then the order of new element will be disrupted after the single element So now new element can be found at `index: 0,2,4,5`
  - But **WHAT IF** array consists of one single element as well? `example:[1,1,2,2,3,4,4]` Then the order of new element will be disrupted after the single element So now new element can be found at `index: 0,2,4,5`
  - Example : `n=7` `arr=[3,3,7,7,10,11,11]` `mid= index 3;` since mid is **odd** and `arr[mid] == arr[mid-1]` it means upto index 3 the array consists of all the duplicates so we move towards right and rellocate low to mid+1;
   
  <details>
  <summary>Sample Code</summary>

    ```cpp
    int singleNonDuplicate(vector<int> &nums) {
        int low, mid, high, prev, next, n = nums.size();
        low = 0;
        high = nums.size() - 1;
        if (nums.size() == 1)
            return nums[0];
        while (low <= high) {
            mid = low + (high - low) / 2;
            prev = (mid + n - 1) % n;
            next = (mid + 1) % n;
  
            /* if(prev element!=curr element & next element != curr element then it is unique) */ 
            if (nums[mid] != nums[prev] && nums[mid] != nums[next]) // 
                return nums[mid];
            else if (mid % 2 == 1 && nums[mid] == nums[prev] || mid % 2 == 0 && nums[mid] == nums[next])
                low = mid + 1;
            else if (mid % 2 == 0 && nums[mid] == nums[prev] || mid % 2 == 1 && nums[mid] == nums[next])
                high = mid - 1;
        }
        return -1;
    }


    ```
 </details>