def peakFinder_1nums n = lennums for i in rangen if i == 0 and numsi numsi+1 return numsi elif i == n-1 and numsi numsi-1 return numsi elif numsi nu

To modify the algorithm to have a time complexity of O(logn), you can use a binary search approach to find the peak element.

Here's the modified code:

def peakFinder_2(nums):
    start = 0
    end = len(nums) - 1
    
    while start < end:
        mid = (start + end) // 2
        
        if nums[mid] > nums[mid+1]:
            end = mid
        else:
            start = mid + 1
    
    return nums[start]

arr = eval(input())
print(peakFinder_2(arr))

In this modified code, we perform binary search by comparing the middle element with its adjacent element. If the middle element is greater than its adjacent element, then we narrow down the search to the left half of the array. Otherwise, we narrow down the search to the right half of the array.

This binary search approach reduces the search space in each iteration, resulting in a time complexity of O(logn).