Find Peak Element

Based on the source, the other question being asked was -
Given an array of positive integers and target, return True if sum of continues array equal to target.
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.

Function
findPeakElement(nums: int[]) → int

Examples
Example 1
nums = [1,2,3,1]
return = 2
3 is a peak element and your function should return the index number 2.
Example 2
nums = [1,2,1,3,5,6,4]
return = 5
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
-2^31 <= nums[i] <= 2^31 - 1
nums[i] != nums[i + 1] for all valid i.

The trick: you don't need sorted data, you need a direction. Compare nums[mid] with nums[mid+1]. If mid is smaller, a peak must exist on the right, because the array eventually drops to negative infinity at the end. So set lo = mid + 1. Otherwise a peak exists at mid or to its left, so set hi = mid. Loop while lo < hi and return lo. The common pitfall is using lo <= hi with hi = mid, which loops forever, or reading nums[mid+1] out of bounds. Adjacent values are never equal, so there's no tie case to handle. For the subarray-sum-to-target question with positive integers, use a sliding window, since growing and shrinking the window moves the sum predictably. If you freeze on the boundary updates during the live OA, StealthCoder is the hedge that gives you the working loop.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.