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# Binary Search

## [Binary Search](https://leetcode.com/problems/binary-search)

Given a **sorted** (in ascending order) integer array `nums` of `n` elements and a `target` value, write a function to search `target` in `nums`. If `target` exists, then return its index, otherwise return `-1`.

\
 **Example 1:**

```

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

```

**Example 2:**

```

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
```

**Note:**

1. You may assume that all elements in `nums` are unique.
2. `n` will be in the range `[1, 10000]`.
3. The value of each element in `nums` will be in the range `[-9999, 9999]`.

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
public:
    //classic O(log n) solution
    int search(vector<int>& nums, int target)
    {
        if(nums.empty())
            return -1;
        //starting index is in the middle of the array
        //manipulate indexies
        int pos_a = 0, pos_b = nums.size()-1;
        while (pos_a <= pos_b)
        {
            size_t mid = (pos_a+pos_b) / 2;
            if(nums[mid] == target)
                return mid;
            else if(nums[mid] < target)
                pos_a = mid + 1;
            else
                pos_b = mid - 1;
        }

        return -1;
    }
};
```