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Arrays: Left Rotation

  • kevinmathis08 + 10 comments

    Yeah index slicing FTW, here was my 1-liner in definition, lol:

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

n, k = map(int, input().strip().split(' '))
a = list(map(int, input().strip().split(' ')))
answer = array_left_rotation(a, n, k);
print(*answer, sep=' ')
    • Lord_krishna + 1 comment

      is that scala?

      • domar + 0 comments

        It's Python 3

    • aniket_vartak + 1 comment

      you dont need to pass n to your function, right..

      • 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 + 2 comments

            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.

            • julianesteban2p1 + 0 comments

              i'd change '''for _ in range(k%len(a)) ''' to be more eficent

        • navi_srm + 0 comments

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

    • ansimionescu + 1 comment

      k -> k%n

      • frostje + 0 comments

        k%n handles any range of shifts, but the problem specifies n < k so you can just use k in this case

    • NuttyMighty + 1 comment

      This logic fails when k>n.

      • madhuvasu7 + 0 comments

        for that, do:

        k = k%n
    • chrislucas + 0 comments

      Nice, only one line was added

    • marwinko19 + 0 comments

      Perfection. Short, simple, & clean.

    • ahmettortumlu25 + 0 comments

      hmm nice....

    • aliie62 + 0 comments

      It's so smart! well done:)

    • toaa_kama2354 + 0 comments

      Can you please explain how (a[k:]+a[:k]) is working?

    • jae_duk_seo + 0 comments

      Damn this is super impressive.