# 852. Peak Index in a Mountain Array

### Approach 1: Return the largest number

```python
class Solution(object):
    def peakIndexInMountainArray(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        return A.index(max(A))
```

The largest number will always be greater than it's neighbours so if just find the largest number and return it's index we will get the peak index

> **Time Complexity:** *O(N),  both max and index are linear runtime functions*
>
> **Space Complexity:** *O(1)*

### Approach 2: Binary Search for the Peak Index

```python
class Solution:
    def peakIndexInMountainArray(self, A):
        """
        https://leetcode.com/problems/peak-index-in-a-mountain-array/description/
        :type A: List[int]
        :rtype: int
        """
        low, high = 0, len(A)-1
        peak = 0

        while low <= high:
            mid = low + (high-low)//2
            if A[mid-1] < A[mid] > A[mid+1]:
                peak = mid
                return peak
            elif A[mid-1] < A[mid]:
                low = mid+1
            else:
                high = mid-1
```

Like any binary search we have the low and high initialized as 0 and the length of the array, and then we try to find not a specific number but a number that satisfies our criteria for peak index

> **Time Complexity:** *O(logN)*
>
> **Space Complexity:** *O(1)*