Leetcode: Find Peak Element

Problem Statement:

A peak element is an element that is strictly greater than its neighbors.

Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

Constraints:

• 1 <= nums.length <= 1000
• -231 <= nums[i] <= 231 - 1
• nums[i] != nums[i + 1] for all valid i.

Solution:

from typing import List

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        # Initialize left and right pointers
        l, r = 0, len(nums) - 1

        # Perform binary search
        while l < r:
            # Calculate the middle index
            m = (l + r) // 2

            # Check if the peak element is on the left side
            if nums[m] > nums[m + 1]:
                r = m
            else:
                # Check if the peak element is on the right side (inclusive)
                l = m + 1

        # At the end of the loop, left pointer points to the peak element
        return l

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