Tags: "leetcode", "binary-search", access_time 1-min read

Peak Index in a Mountain Array

Created: September 23, 2019 by [lek-tin]

Last updated: September 23, 2019

Let’s call an array A a mountain if the following properties hold:

1. A.length >= 3
2. There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1

Input: [0,1,0]
Output: 1

Example 2

Input: [0,2,1,0]
Output: 1

Note

1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.

Solution

class Solution:
    def peakIndexInMountainArray(self, nums: List[int]) -> int:
        if not nums or len(nums) < 3:
            return -1

        start, end = 0, len(nums)-1

        while start + 1 < end:
            mid = start + (end-start)//2
            if nums[mid] < nums[mid-1]:
                end = mid
            elif nums[mid] < nums[mid+1]:
                start = mid
            else:
                return mid