The reader is advised to read all the lessons (tutorials) in this full course, in the order presented.

Problem

Given an array, arr[] and an integer k=3, find the array after reversing every sub-array of consecutive k elements in place. If the last sub-array has fewer than k elements, reverse it as it is. Modify the array in place; do not return anything. The given array is:

    [0, 1, 2, 3, 4, 5, 6, 7]  

The algorithm should use O(n) time and O(n) space. The output should be:

    [2, 1, 0, 5, 4, 3, 7, 6]  

Solution

Edge Cases:

When k = 1, or k = 0, the array stays the same.
When k is greater than or equal to the array size, the whole array is reversed. 

Strategy

- Take the first edge case into consideration.
- Begin from index 0 and find the size of the current sub-array to be reversed. If the number of elements is less than k, reverse all of them.
- Each sub-array is reversed using two pointers (optionally) that start from the two corners of the sub-array.

In the function there is the principal for-loop, where increment of the index is done in k's.

Zero based indexing is used. The left included index of a group (sub-array) is obtained from,

    int left = i;

where i is the iterating index. The right included index of a group (sub-array) is obtained from,

    i+k-1    //zero based counting

where i is the iterating index (zero based).

The following program illustrates this (read through the code and comments):

def reverse_array_in_groups(arr, k)
  n = arr.length
  
  i = 0
  while i < n
    left = i
    # to determine right value
    right = (i + k - 1 < n - 1) ? i + k - 1 : n - 1    #zero based indexing

    # reverse the sub-array [left, right]
    while left < right
      # swap
      temp = arr[left]
      arr[left] = arr[right]
      arr[right] = temp
      
      left += 1
      right -= 1
    end
    
    i += k
  end
end

arr = [0, 1, 2, 3, 4, 5, 6, 7]
k = 3
reverse_array_in_groups(arr, k)

for i in 0...arr.length
  print(arr[i], " ")
end
puts

The output is:

    2 1 0 5 4 3 7 6 

Thanks for reading.