Find peak element leetcode 162 python

certainly! the "find peak element" problem is a classic problem often encountered in coding interviews. it is part of leetcode problem number 162. the problem statement is as follows:

## problem statement

you are given an integer array `nums` where `nums[i] != nums[i + 1]`, and you need to find a peak element and return its index. a peak element is defined as an element that is greater than its neighbors. for corner elements, you only need to consider one neighbor.

### key points
- an element `nums[i]` is a peak if:
- if `i` == 0, then `nums[i]` should be greater than `nums[i + 1]`.
- if `i` == `n-1`, then `nums[i]` should be greater than `nums[i - 1]`.
- for other elements, `nums[i]` should be greater than both `nums[i - 1]` and `nums[i + 1]`.

### constraints
- the array will not have duplicate elements.
- the array size will be at least 1.

### example

## approach
we can solve this problem using a binary search approach, which allows us to find a peak in o(log n) time. the idea is to leverage the property of peaks in the array. if the middle element is not a peak, we can decide which half of the array to explore further based on the values of its neighbors.

### binary search steps:
1. start with two pointers: `left` at 0 and `right` at `n-1` (where `n` is the length of the array).
2. calculate the middle index `mid = (left + right) // 2`.
3. check if `nums[mid]` is greater than its neighbors:
- if yes, `nums[mid]` is a peak, return `mid`.
- if not, check the neighbors:
- if `nums[mid] nums[mid + 1]`, it means there is a peak on the right side, so move `left` to `mid + 1`.
- if `nums[mid] nums[mid - 1]`, it indicates a peak might be on the left side, so move `right` to `mid - 1`.
4. repeat this process until you find a peak.

### python code example

here’s a python implementation of the above approach:

### explanation of the code:
- we define a function `findpeakelement` that takes a list of integers `nums`.
- we initialize `lef ...

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