## Description

https://leetcode.com/problems/find-peak-element/

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

Given an integer array `nums`

, find a peak element, and return its index. If the array contains multiple peaks, return the index to **any of the peaks**.

You may imagine that `nums[-1] = nums[n] = -∞`

.

You must write an algorithm that runs in `O(log n)`

time.

**Example 1:**

Input:nums = [1,2,3,1]Output:2Explanation:3 is a peak element and your function should return the index number 2.

**Example 2:**

Input:nums = [1,2,1,3,5,6,4]Output:5Explanation:Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

**Constraints:**

`1 <= nums.length <= 1000`

`-2`

^{31}<= nums[i] <= 2^{31}- 1`nums[i] != nums[i + 1]`

for all valid`i`

.

## Explanation

Use binary search to find the peak element which is greater than the nearby elements.

## Python Solution

```
class Solution:
def findPeakElement(self, nums: List[int]) -> int:
start = 0
end = len(nums) - 1
while start + 1 < end:
mid = start + (end - start) // 2
if nums[mid] > nums[mid - 1] and nums[mid] > nums[mid + 1]:
return mid
elif nums[mid] < nums[mid - 1]:
end = mid
elif nums[mid] < nums[mid + 1]:
start = mid
if nums[end] > nums[start]:
return end
return start
```

- Time complexity: O(log(N)).
- Space complexity: O(1).

I found that solution is very popular and helpful: https://youtu.be/U9iw_aTuGV0