'''
Let's call an array arr a mountain if the following properties hold:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... arr[i-1] < arr[i]
arr[i] > arr[i+1] > ... > arr[arr.length - 1]
Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].
Example 1:
Input: arr = [0,1,0]
Output: 1

Example 3:
Input: arr = [0,10,5,2]
Output: 1

Example 4:
Input: arr = [3,4,5,1]
Output: 2

Example 5:
Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2
 
Constraints:
0 <= arr[i] <= 106
arr is guaranteed to be a mountain array.
'''

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        
        # lazy solution:
        # return arr.index(max(arr))
        
        # binary search solution
        left, right = 0, len(arr) - 1
        while left < right:
            mid = (left + right) // 2
            if arr[mid - 1] <= arr[mid] and arr[mid] >= arr[mid + 1]:
                return mid
            elif arr[mid] < arr[mid + 1]:
                left = mid
            else:
                right = mid

'''
Let's call an array arr a mountain if the following properties hold:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... arr[i-1] < arr[i]
arr[i] > arr[i+1] > ... > arr[arr.length - 1]
Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].
Example 1:
Input: arr = [0,1,0]
Output: 1

Example 3:
Input: arr = [0,10,5,2]
Output: 1

Example 4:
Input: arr = [3,4,5,1]
Output: 2

Example 5:
Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2
 
Constraints:
0 <= arr[i] <= 106
arr is guaranteed to be a mountain array.
'''

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        
        # lazy solution:
        # return arr.index(max(arr))
        
        # binary search solution
        left, right = 0, len(arr) - 1
        while left < right:
            mid = (left + right) // 2
            if arr[mid - 1] <= arr[mid] and arr[mid] >= arr[mid + 1]:
                return mid
            elif arr[mid] < arr[mid + 1]:
                left = mid
            else:
                right = mid

Let's call an array arr a mountain if the following properties hold:  arr.length >= 3 There exists some i with 0 < i < arr.length - 1 such that: arr[0] < arr[1] < ... arr[i-1] < arr[i] arr[i] > arr[i+1] > ... > arr[arr.length - 1] Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1]. Example 1: Input: arr = [0,1,0] Output: 1  Example 3: Input: arr = [0,10,5,2] Output: 1  Example 4: Input: arr = [3,4,5,1] Output: 2  Example 5: Input: arr = [24,69,100,99,79,78,67,36,26,19] Output: 2   Constraints: 0 <= arr[i] <= 106 arr is guaranteed to be a mountain array. # lazy solution: # return arr.index(max(arr)) # binary search solution