### Problem Statement

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.

**Example 1:**

```
Input: [1,1,2,3,3,4,4,8,8]
Output: 2
```

**Example 2:**

```
Input: [3,3,7,7,10,11,11]
Output: 10
```

**Note:** Your solution should run in O(log n) time and O(1) space.

### Solution

```
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int L=0, U=nums.size()-1;
int result;
while(L<=U)
{
if(L==U)
{
result = nums[L];
break;
}
int M = L+(U-L)/2;
int consideredLength = M-L+1;
if(consideredLength%2 == 1)
{
if(M-1>=0 && nums[M] == nums[M-1]) {
U = M-2;
}
else if(M+1<= nums.size()-1 && nums[M] == nums[M+1]) {
L = M+2;
}
else {
result = nums[M];
break;
}
}
else {
if(M+1<= nums.size()-1 && nums[M] == nums[M+1]) {
U = M-1;
}
else if(M-1>=0 && nums[M] == nums[M-1]) {
L = M+1;
}
else {
result = nums[M];
break;
}
}
}
return result;
}
};
```

### Solution Thought Process

I solved this problem using binary search. First, we get the middle element of the range using low and high.

We get the length considered by calculating:

```
consideredLength = M-L+1
```

- Let's consider if this length is odd.

```
nums 1 1 2 2 3
idx 0 1 2 3 4
L = 0, U = 4
M = 0 + (4-0)/2 = 2
```

So`[L, M]`

is odd in length. If this is the case, if `nums[M] == nums[M+1]`

, it means that the element can be found from element index `M+2`

. So we make `L=M+2`

Let's see another case,

```
nums 1 2 2 3 3
idx 0 1 2 3 4
L = 0, U = 4
M = 0 + (4-0)/2 = 2
```

So `[L, M]`

is odd in length. If `nums[M] == nums[M-1]`

, it means that the element can be found before element index `M-2`

, included.

If those are not the case, then `nums[M]`

is the result and we break the loop.

- Let's consider if the considered length is even.

```
nums 1 1 2 3 3 5 5
idx 0 1 2 3 4 5 6
L = 0, U = 6
M = 0 + (6-0)/2 = 3
```

So `[L, M]`

is even in length. If `nums[M] == nums[M+1]`

, it means that the element can be found before element index `M-1`

, included . So we make `U = M-1`

```
nums 1 1 2 2 3 5 5
idx 0 1 2 3 4 5 6
L = 0, U = 6
M = 0 + (6-0)/2 = 3
```

So `[L, M]`

is even in length. If `nums[M]==nums[M-1]`

, it means that the element can be found from element index `M+1`

. So we make `L=M+1`

.

If we have got `L`

and `U`

as the same element, we return the element as the result.

### Complexity

**Time Complexity:** O(logn), we are making the consideration space half in every iteration.

**Space Complexity:** O(1)

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