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):

package main

import (
	"fmt"
)

    func reverseArrayInGroups(arr []int, k int) { 
        var n = len(arr);
    
        for i := 0; i < n; i += k {
            var left = i;
            var right int;
    
            // 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]
            for (left < right) {
                // swap
                var temp = arr[left];
                arr[left] = arr[right];
                arr[right] = temp;

                left++;
                right--;
            }
        }
    }

func main() {
    var arr = []int{0, 1, 2, 3, 4, 5, 6, 7};
    var k = 3;
        
    reverseArrayInGroups(arr, k);
        
    for i := 0; i < len(arr); i++ { 
        fmt.Print(arr[i], " ");
    }
    fmt.Println();
}

The output is:

    2 1 0 5 4 3 7 6 

Thanks for reading.