rotate_with_cycles_en.py

def rotate_with_cycles(arr,k): #inputs are array and number of steps to rotate k
    if k == 0:
        return arr #corner case
    n = len(arr)
    moved_counter = 0 #additional memory (total O(1))
    cur_start = 0
    pointer = k
    prev_element = arr[0]
    while moved_counter <n:
        buf = arr[pointer] #one more variable O(1) memory
        arr[pointer] = prev_element
        prev_element = buf
        moved_counter += 1
        if pointer == cur_start: #if we meet a cycle we possibly need to move to the right once
            cur_start += 1
            pointer = cur_start
            prev_element = arr[cur_start]
        pointer = (pointer + k) % n
    #total memory O(1)
    #time complexity O(n) since we traverse each element no more than once
    return arr

print(rotate_with_cycles(list(range(10)),3))
#Output: [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]