> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/binary-search/template-iii/3.find-peak-element.md).

# 3.Find Peak Element

A peak element is an element that is greater than its neighbors.

**Algorithm:**

1. Check if the **mid value** or index mid = (l+r)/2, is the **peak element or not**, if yes then print the element and terminate. (check of if both the left and right neighbors are greater)
2. Else if the **element on the left side of the middle element is greater** then check for **peak element on the left side**, i.e. update r = mid – 1
3. Else if the **element on the right side of the middle element is greater** then check for **peak element on the right side ,** i.e. update l = mid + 1

#### Corner Cases: 

1. If **input array is sorted in strictly increasing order**, the **last element** is always a **peak element**. For example, 50 is peak element in {10, 20, 30, 40, 50}.
2. If the **input array is sorted in strictly decreasing order**, the **first element** is always a **peak element.** 100 is the peak element in {100, 80, 60, 50, 20}.

## Solution : (Binary Search)

```cpp
class Solution
{
public:
    int findPeakElement(vector<int> &nums)
    {

        if (nums.size() == 1)
        {
            return 0;
        }

        int start = 0;
        int end = nums.size() - 1;
        int res = -1;
        while (start <= end)
        {

            int mid = (start + end) / 2;

            if (mid > 0 && mid < nums.size() - 1)
            {

                if (nums[mid] > nums[mid + 1] && nums[mid] > nums[mid - 1])
                {
                    res = mid;
                    break;
                }
                else if (nums[mid] < nums[mid + 1])
                {
                    start = mid + 1;
                }
                else
                {
                    end = mid - 1;
                }
            }
            
            else if (mid == 0)
            {
                if (nums[mid] > nums[mid + 1])
                {
                    res = mid;
                    break;
                }
                else
                {
                    start = mid + 1;
                }
            }

            else if (mid == nums.size() - 1)
            {
                if (nums[mid] > nums[mid - 1])
                {
                    res = mid;
                    break;
                }
                else
                {
                    end = mid - 1;
                }
            }
        }

        return res;
    }
};
```

**Time Complexity: O(log (n))**