# Solving Leetcode Interviews in Seconds with AI: Peak Index in a Mountain Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "852" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an integer mountain array arr of length n where the values increase to a peak element and then decrease. Return the index of the peak element. Your task is to solve it in O(log(n)) time complexity.   Example 1:  Input: arr = [0,1,0] Output: 1  Example 2:  Input: arr = [0,2,1,0] Output: 1  Example 3:  Input: arr = [0,10,5,2] Output: 1    Constraints:  3 <= arr.length <= 105 0 <= arr[i] <= 106 arr is guaranteed to be a mountain array.  

	# Explanation
	Here's the approach to find the peak element in a mountain array efficiently:

*   **Binary Search:** Utilize binary search to narrow down the search space by repeatedly dividing the array into halves.
*   **Peak Identification:** In each iteration, compare the middle element with its neighbors to determine if it's the peak or if the peak lies in the left or right half.
*   **Termination:** Continue the binary search until the peak element is found.

*   **Runtime Complexity:** O(log n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def peakIndexInMountainArray(arr):
    """
    Finds the index of the peak element in a mountain array.

    Args:
        arr: A list of integers representing the mountain array.

    Returns:
        The index of the peak element.
    """
    left, right = 0, len(arr) - 1

    while left < right:
        mid = (left + right) // 2

        if arr[mid] < arr[mid + 1]:
            left = mid + 1  # Peak is to the right
        else:
            right = mid  # Peak is mid or to the left

    return left
	```