LC852. Peak Index in a Mountain Array

Problem Description

LeetCode Problem 852: An array arr is 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 a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

You must solve it in O(log(arr.length)) time complexity.

Clarification

Assumption

Solution

Approach - Binary Search

We can use binary search to find the peak by checking whether arr[mid] < arr[mid + 1]:

• true, means climbing up from left, the peak is on the right half
• false, means on the right side, peak is on the left half

Python

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        left, right = 0, len(arr) - 1

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

            if arr[mid] < arr[mid + 1]:
                left = mid + 1
            else:
                right = mid

        return left

Complexity Analysis

• Time complexity: \(O(\log n)\)
  Since using binary search, the time complexity is \(O(\log n)\).
• Space complexity: \(O(1)\)
  Only use two index variables.

Test