Let's call an array arr
a mountain if the following properties hold:
-
arr.length >= 3
- There exists some
i
with0 < 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 2:
Input: arr = [0,2,1,0]
Output: 1
Example 3:
Input: arr = [0,10,5,2]
Output: 1
Constraints:
-
3 <= arr.length <= 104
-
0 <= arr[i] <= 106
-
arr
is guaranteed to be a mountain array.
Follow up: Finding the O(n)
is straightforward, could you find an O(log(n))
solution?SOLUTION:
class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
old = True
new = True
for i in range(len(arr) - 1):
new = arr[i + 1] > arr[i]
if new ^ old:
return i
old = new
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