Tells whether max 1 swap is needed to sort an array in Python

swaps_needed.py

from itertools import tee, izip
from copy import copy


def pairwise(seq):
    a, b = tee(seq)
    b.next()
    return izip(a, b)


def solution(A):
    # O(N log(N)) but I think we can do better
    ordered = True
    for current_item, next_item in pairwise(A):
        if current_item > next_item:
            ordered = False
            break
    if ordered:
        return True
    unsorted_array = copy(A)
    A.sort()  # timsort takes O(n logn)
    count = 0
    for n, elem in enumerate(A):
        if unsorted_array[n] != elem:
            if count < 2:
                count += 1
            else:
                return False
    return True

print solution([1, 5, 3, 3, 7])  # returns True, swap one (5,3) to get sorted array
print solution([1, 3, 3, 3, 7])  # returns True because list is already sorted
print solution([1, 3, 5, 3, 4])  # returns False, 2 swaps are needed