Find the monotonic pair whose indices are the furthest apart in Python

distance_pairs.py

# method helps with fact that enumerate from reversed
# won't yield reversed indexes
def reverse_enum(L):
    # guaranteed to NOT make any extra copy or search operations, O(1)
    for index in reversed(xrange(len(L))):
        yield index, L[index]


def solution(A):
    top = list()
    maximum = float("-inf")
    # allocate temp list with O(n) time, put there maximal values descending
    for item in reversed(A):
        if (item > maximum):
            maximum = item
        top.insert(0, maximum)
    best = 0
    cur_max_index = 0
    # iterate with O(n) time
    for i, item in enumerate(A):
        # simply calculate the distance, takes O(c*n) time, c between 0 and 1
        while(cur_max_index < len(top) and top[cur_max_index] >= item):
            cur_max_index += 1
        if((cur_max_index - 1 - i) > best):
            best = cur_max_index - 1 - i
    return best


print solution([0])
print solution([0, 0])
print solution([1, 3, -3])
print solution([5, 3, 6, 3, 4, 2])

Inspired by: stackoverflow.com - monotonic-pair-codility