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  1. Practice
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  4. Arrays: Left Rotation
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Arrays: Left Rotation

  • michael_bubb + 2 comments

    I agree - I ended up not using 'n' (Python):

    def left_shift(n,k,a):
   for _ in range(k):
      a.append(a.pop(0))
   print(*a)
    • unitraxx + 1 comment

      Obviously this solves the problem, but is a terrible solution. Pop is an O(N) operation. So your solution becomes O(K*N). This should be done in O(N) total time complexity. You do have the space requirement of O(1) correct. All the standard solutions have a O(N) space complexity.

      • asfaltboy + 1 comment

        True. However, it becomes an elegant solution if we use collections.dequeue instead of list. Double ended queues have a popleft method, which is an O(1) operation:

        def array_left_rotation(a, n, k):
    for _ in range(k):
        a.append(a.popleft())
    return a

        More info: https://wiki.python.org/moin/TimeComplexity#collections.deque https://docs.python.org/3/library/collections.html#collections.deque

        • AffineStructure + 1 comment

          They have rotate built into the deque

          def array_left_rotation(a, n, k):
    a = deque(a)
    for i in range(k):
        a.rotate(-1)
    return a
          • josegabriel_st + 0 comments

            You have O(n) in:

            a = deque(a)

            In order to avoid this, you should use a deque from the beginning like:

            from collections import deque

def array_left_rotation(a, n, k):
    a.rotate(-k)

n, k = map(int, input().strip().split(' '))
a = deque(map(int, input().strip().split(' ')))
array_left_rotation(a, n, k);
print(*a, sep=' ')

            And array_left_rotation only takes O(k) instead of O(n).

            Note that a is pased by reference, so there is no need to return anything, but this could be an issue for some user cases, for this particular problem, it works.

    • navi_srm + 0 comments

      def array_left_rotation(a, n, k): return (a[k:] + a[:k])