#cp-medium 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.

My Solution

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        L = 0
        R = len(nums) -1

        while (L<=R):
            mid = L + (R-L)//2
            print(mid,L,R)
            if len(nums)==1:
                return 0
            
            
            if mid>=1 and mid <= len(nums)-2:
                if nums[mid]>nums[mid-1] and nums[mid]>nums[mid+1]:
                    return mid
            elif mid == 0 and  nums[mid]>nums[mid+1]:
                return mid
            elif mid == len(nums)-1 and nums[mid]>nums[mid-1]:
                return mid

            if (mid!=0 and nums[mid-1]>nums[mid]) or mid==len(nums)-1:
                R = mid -1
            elif (mid!=(len(nums)-1) and nums[mid+1]>nums[mid]) or mid==0:
                L = mid + 1
        return -1
        wayyy too mach conditions

Best Solution

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

        while left < right:
            mid = (left + right) // 2
            if nums[mid] > nums[mid + 1]:
                right = mid
            else:
                left = mid + 1

        return left

How to Improve my Solution?

The issue isn’t that you missed edge cases — it’s that you picked an approach that forces you to handle edge cases manually.

Why your solution explodes with conditions

You’re asking: “is mid a peak?” at every step. To answer that, you need to check both neighbors, which means you need to guard against mid=0 and mid=n-1 separately. That’s where all the noise comes from.

Why the clean solution has zero edge cases

It never asks “is this a peak?”. It only asks one thing:

Which side is going uphill?

if nums[mid] > nums[mid + 1]:
    right = mid   # peak is on the left side (could be mid itself)
else:
    left = mid + 1  # nums[mid+1] > nums[mid], so peak is strictly to the right

mid + 1 is always safe because the loop condition is left < right, which guarantees mid < right <= n-1, so mid+1 always exists. No boundary checks needed at all.

The insight that makes it work: a peak must exist on the uphill side. If nums[mid+1] > nums[mid], the right side is climbing — a peak must exist there (even if the array just keeps climbing, the last element is a peak since nums[n] = -∞).