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 reverseArrayInGroups(arr, k):
    n = len(arr);
    
    for i in range(0, n, k):
        left = i;
        right = n;
    
        #to determine right value
        if i+k-1 < n-1:
            right = i+k-1;    #zero based indexing
        else:
            right = 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;

#main area
arr = [0, 1, 2, 3, 4, 5, 6, 7];
k = 3;
        
reverseArrayInGroups(arr, k);
        
for i in range(0, len(arr)):
    print(arr[i], end=" ");
print();

The output is:

    2 1 0 5 4 3 7 6 

Thanks for reading.