1. Problem Formalization & Boundary Conditions

Let $A = [a_0, a_1, \dots, a_{n-1}]$ be an array of $n$ distinct adjacent integers. An index $i \in \{0, \dots, n-1\}$ is a peak element if and only if:

• $a_i > a_{i-1}$ (where $a_{-1} = -\infty$)
• $a_i > a_{i+1}$ (where $a_n = -\infty$)

These boundary conditions produce two vital structural consequences:

1. Boundary Peaks: The first element $a_0$ is a valid peak if $a_0 > a_1$. Similarly, the final element $a_{n-1}$ is a valid peak if $a_{n-1} > a_{n-2}$. In a single-element array ($n = 1$), index 0 is always a valid peak because both virtual neighbors are $-\infty$.
2. Guaranteed Existence: Because adjacent values are never equal and both boundaries drop to $-\infty$, the sequence cannot be infinitely flat. By the discrete extreme value theorem on finite sequences, every valid input array is guaranteed to contain at least one local peak element.

2. Binary Search Convergence Proof on Unsorted Arrays

The naive approach is a linear scan comparing each element to its neighbors, requiring $O(n)$ time. However, the strict problem constraint demands an $O(\log n)$ runtime. How can binary search function without sorted data?

The key insight lies in local slope invariant preservation. Consider the midpoint index mid and its adjacent right neighbor mid + 1:

• Case 1: Ascending Slope (nums[mid] < nums[mid + 1]): The array is currently rising towards the right. Since the far right boundary drops off to $-\infty$ at index $n$, the sequence cannot increase indefinitely. It must either continue rising until reaching the boundary element $a_{n-1}$ (making $a_{n-1}$ a peak) or drop at some intermediate point (creating an internal peak). Therefore, a peak is mathematically guaranteed to exist in the right subarray $[mid + 1, \text{right}]$. We discard the left half: left = mid + 1.
• Case 2: Descending Slope (nums[mid] > nums[mid + 1]): The array is falling towards the right. This means index mid itself could be a peak (if it is also greater than mid - 1), or an upward peak occurred earlier to the left. Since the far left boundary drops to $-\infty$ at index $-1$, a peak is guaranteed to exist in the left subarray $[\text{left}, mid]$. We discard the right half: right = mid.

3. Production Multi-Language Implementations

Here are production-grade implementations demonstrating clean loop termination, integer overflow protection during midpoint calculation, and strict typing.

C++20 Implementation

#include <vector>
#include <cstddef>

class Solution {
public:
    int findPeakElement(const std::vector<int>& nums) {
        int left = 0;
        int right = static_cast<int>(nums.size()) - 1;

        // Invariant: Peak is guaranteed within [left, right]
        while (left < right) {
            // Prevent integer overflow: left + (right - left) / 2
            int mid = left + (right - left) / 2;

            if (nums[mid] < nums[mid + 1]) {
                // Ascending slope: peak must be to the right
                left = mid + 1;
            } else {
                // Descending slope: mid itself or an element to the left is peak
                right = mid;
            }
        }

        // Loop terminates when left == right, pointing directly at a peak index
        return left;
    }
};

TypeScript Implementation

/**
 * Finds the index of any local peak element in O(log n) time.
 * @param nums - Array of integers where nums[i] !== nums[i+1]
 * @returns Index of a confirmed peak element
 */
export function findPeakElement(nums: number[]): number {
  let left = 0;
  let right = nums.length - 1;

  while (left < right) {
    const mid = Math.floor(left + (right - left) / 2);

    if (nums[mid] < nums[mid + 1]) {
      // Ascending gradient: move search space to right sub-interval
      left = mid + 1;
    } else {
      // Descending gradient: peak lies at mid or to the left
      right = mid;
    }
  }

  return left;
}

Python 3 Implementation

from typing import List

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        
        while left < right:
            mid = left + (right - left) // 2
            
            if nums[mid] < nums[mid + 1]:
                left = mid + 1
            else:
                right = mid
                
        return left

4. Asymptotic Complexity & Edge Cases

5. Key Architectural Takeaways for Interview Candidates

In technical interview environments (FAANG/Tier-1 engineering rounds), interviewers select this problem to evaluate candidates' understanding of search invariants:

• Binary Search Does Not Require Global Sorting: It requires only a monotonic decision boundary or a guaranteed path to a solution. As long as you can discard one half of the search space with absolute certainty, binary search is mathematically valid.
• Strictly Handle Edge Invariants: The condition while (left < right) combined with right = mid avoids infinite loops. When only two elements remain (e.g. left = 0, right = 1), mid = 0, and mid + 1 = 1 is safely in bounds. If nums[0] < nums[1], left = 1 and loop exits. If nums[0] > nums[1], right = 0 and loop exits.
• Multiple Peaks Permitted: The problem specification allows returning any peak. You do not need to locate the global maximum; finding any local extremum satisfies all problem constraints.