Reverse an array in groups of given size | Set 2 (Variations of Set 1 )

Given an array, reverse every sub-array that satisfies the given constraints.
We have discussed a solution where we reverse every sub-array formed by consecutive k elements in Set 1. In this set, we will discuss various interesting variations of this problem. 

Variation 1 (Reverse Alternate Groups): Reverse every alternate sub-array formed by consecutive k elements.

Examples:

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
k = 3
Output:  
[3, 2, 1, 4, 5, 6, 9, 8, 7]

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 2
Output:  
[2, 1, 3, 4, 6, 5, 7, 8]

Algorithm:

1. Define a function reverse() that takes an integer array arr, its size n, and the size of the sub-arrays k as input
2. Traverse the array in multiples of 2k starting from the first element, i.e., for i = 0, 2k, 4k, and so on.
3. For each such i, set the left pointer to i and the right pointer to min(i + k – 1, n – 1) to handle the case when 2k is not a multiple of n.
4. Reverse the sub-array [left, right] using a while loop and the swap() function.
5. Repeat steps 3-4 for every alternate sub-array formed by consecutive k elements.
6. In the main function, declare an integer array arr, initialize it with some values, and define the size of the sub-arrays k.
7. Determine the size of the array n using the sizeof() operator.
8. Call the reverse() function passing the integer array arr, its size n, and the size of the sub-arrays k as input.
9. Print the modified array arr.

Below is the implementation :

C++

// C++ program to reverse every alternate sub-array
// formed by consecutive k elements
#include <iostream>
using namespace std;

// Function to reverse every alternate sub-array
// formed by consecutive k elements
void reverse(int arr[], int n, int k)
{
    // increment i in multiples of 2*k
    for (int i = 0; i < n; i += 2*k)
    {
        int left = i;

        // to handle case when 2*k is not multiple of n
        int right = min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr[left++], arr[right--]);
    }
}

// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14};
    int k = 3;

    int n = sizeof(arr) / sizeof(arr[0]);

    reverse(arr, n, k);

    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Java

// Java program to reverse
// every alternate sub-array
// formed by consecutive k elements
class GFG
{

// Function to reverse every
// alternate sub-array formed
// by consecutive k elements
    static void reverse(int arr[], int n, int k)
    {

        // increment i in multiples of 2*k
        for (int i = 0; i < n; i += 2 * k)
        {
            int left = i;

            // to handle case when 2*k is not multiple of n
            int right = Math.min(i + k - 1, n - 1);

            // reverse the sub-array [left, right]
            while (left < right) {
                swap(arr, left++, right--);
            }
        }
    }

    static int[] swap(int[] array, int i, int j)
    {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
        return array;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5, 6, 7, 8,
                    9, 10, 11, 12, 13, 14};
        int k = 3;
        int n = arr.length;

        reverse(arr, n, k);
        for (int i = 0; i < n; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to reverse every alternate sub-array
# formed by consecutive k elements

# Function to reverse every alternate sub-array
# formed by consecutive k elements
def reverse(arr, n, k):
    # increment i in multiples of 2*k
    for i in range(0,n,2*k):
        left = i

        # to handle case when 2*k is not multiple of n
        right = min(i + k - 1, n - 1)

        # reverse the sub-array [left, right]
        while (left < right):
            temp = arr[left]
            arr[left] = arr[right]
            arr[right] = temp
            left += 1
            right -= 1

# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
    k = 3

    n = len(arr)

    reverse(arr, n, k)

    for i in range(0,n,1):
        print(arr[i],end =" ")

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to reverse every alternate
// sub-array formed by consecutive k elements
using System;

class GFG
{

    // Function to reverse every
    // alternate sub-array formed
    // by consecutive k elements
    static void reverse(int []arr,
                        int n, int k)
    {

        // increment i in multiples of 2*k
        for (int i = 0; i < n; i += 2 * k)
        {
            int left = i;

            // to handle case when 2*k is
            // not multiple of n
            int right = Math.Min(i + k - 1, n - 1);

            // reverse the sub-array [left, right]
            while (left < right)
            {
                swap(arr, left++, right--);
            }
        }
    }

    static int[] swap(int[] array, int i, int j)
    {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
        return array;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7, 8,
                     9, 10, 11, 12, 13, 14};
        int k = 3;
        int n = arr.Length;

        reverse(arr, n, k);
        for (int i = 0; i < n; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to reverse
// every alternate sub-array
// formed by consecutive k elements

// Function to reverse every
// alternate sub-array formed
// by consecutive k elements
function reverse(arr, n, k)
{

    // Increment i in multiples of 2*k
    for(let i = 0; i < n; i += 2 * k)
    {

        let left = i;

        // To handle case when 2*k is
        // not multiple of n
        let right = Math.min(i + k - 1,
                                 n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
        {
            swap(arr, left++, right--);
        }
    }
}

function swap(array, i, j)
{
    let temp = array[i];
    array[i] = array[j];
    array[j] = temp;
    return array;
}

// Driver code
let arr = [ 1, 2, 3, 4, 5, 6, 7, 8,
            9, 10, 11, 12, 13, 14 ];
let k = 3;
let n = arr.length;

reverse(arr, n, k);

for(let i = 0; i < n; i++)
{
    document.write(arr[i] + " ");
}

// This code is contributed by rag2127

</script>

Output

3 2 1 4 5 6 9 8 7 10 11 12 14 13 

Time Complexity: O(N)

Auxiliary Space: O(1)

Variation 2 (Reverse at given distance): Reverse every sub-array formed by consecutive k elements at given distance apart. 

Examples: 

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
k = 3
m = 2
Output:  
[3, 2, 1, 4, 5, 8, 7, 6, 9, 10]

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
k = 3
m = 1
Output:  
[3, 2, 1, 4, 7, 6, 5, 8, 10, 9]

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 2
m = 0
Output:  
[2, 1, 4, 3, 6, 5, 8, 7]

Below is its implementation:

C++

// C++ program to reverse every sub-array formed by
// consecutive k elements at given distance apart
#include <iostream>
using namespace std;

// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
void reverse(int arr[], int n, int k, int m)
{
    // increment i in multiples of k + m
    for (int i = 0; i < n; i += k + m)
    {
        int left = i;

        // to handle case when k + m is not multiple of n
        int right = min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr[left++], arr[right--]);
    }
}

// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14};
    int k = 3;
    int m = 2;

    int n = sizeof(arr) / sizeof(arr[0]);

    reverse(arr, n, k, m);

    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Java

// java program to reverse every sub-array formed by
// consecutive k elements at given distance apart
class GFG
{

// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
static void reverse(int[] arr, int n, int k, int m)
{
    // increment i in multiples of k + m
    for (int i = 0; i < n; i += k + m)
    {
        int left = i;

        // to handle case when k + m is not multiple of n
        int right = Math.min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr,left++, right--);
    }
}
 static int[] swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }

// Driver code
 public static void main(String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5, 6, 7, 8,
                    9, 10, 11, 12, 13, 14};
        int k = 3;
        int m = 2;
        int n = arr.length;

        reverse(arr, n, k, m );
        for (int i = 0; i < n; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }
}
// This code has been contributed by Rajput-Ji

Python3

# Python3 program to reverse every
# sub-array formed by consecutive
# k elements at given distance apart

# Function to reverse every
# sub-array formed by consecutive
# k elements at m distance apart
def reverse(arr, n, k, m):

    # increment i in multiples of k + m
    for i in range(0, n, k + m):
        left = i;

        # to handle case when k + m
        # is not multiple of n
        right = min(i + k - 1, n - 1);

        # reverse the sub-array [left, right]
        while (left < right):
            arr = swap(arr,left, right);
            left += 1;
            right -= 1;
    return arr;

def swap(arr, i, j):

    temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;

# Driver code
arr = [1, 2, 3, 4, 5, 6, 7,
       8, 9, 10, 11, 12, 13, 14];
k = 3;
m = 2;
n = len(arr);

arr = reverse(arr, n, k, m );
for i in range(0, n):
    print(arr[i], end = " ");

# This code is contributed by Rajput-Ji

C#

// C# program to reverse every sub-array
// formed by consecutive k elements at
// given distance apart
using System;

class GFG
{

// Function to reverse every sub-array
// formed by consecutive k elements
// at m distance apart
static void reverse(int[] arr, int n,
                    int k, int m)
{
    // increment i in multiples of k + m
    for (int i = 0; i < n; i += k + m)
    {
        int left = i;

        // to handle case when k + m is
        // not multiple of n
        int right = Math.Min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr, left++, right--);
    }
}

static int[] swap(int[] arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = {1, 2, 3, 4, 5, 6, 7, 8,
                 9, 10, 11, 12, 13, 14};
    int k = 3;
    int m = 2;
    int n = arr.Length;

    reverse(arr, n, k, m );
    for (int i = 0; i < n; i++)
    {
        Console.Write(arr[i] + " ");
    }
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// javascript program to reverse every sub-array formed by
// consecutive k elements at given distance apart

// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
function reverse(arr,n,k,m)
{
    // increment i in multiples of k + m
    for (let i = 0; i < n; i += k + m)
    {
        let left = i;

        // to handle case when k + m is not multiple of n
        let right = Math.min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr,left++, right--);
    }
}

function swap(arr,i,j)
{
    let temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
}

// Driver code
let arr=[1, 2, 3, 4, 5, 6, 7, 8,
                    9, 10, 11, 12, 13, 14];
let k = 3;
let m = 2;
let n = arr.length;

reverse(arr, n, k, m );
for (let i = 0; i < n; i++)
{
    document.write(arr[i] + " ");
}

// This code is contributed by ab2127
</script>

Output

3 2 1 4 5 8 7 6 9 10 13 12 11 14 

Time Complexity: O(N)

Auxiliary Space: O(1)

Variation 3 (Reverse by doubling the group size):
Reverse every sub-array formed by consecutive k elements where k doubles itself with every sub-array.

Examples:

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
k = 1
Output:  
[1], [3, 2], [7, 6, 5, 4], [15, 14, 13, 12, 11, 10, 9, 8]

Below is its implementation:

C++

// C++ program to reverse every sub-array formed by
// consecutive k elements where k doubles itself with
// every sub-array.
#include <iostream>
using namespace std;

// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
void reverse(int arr[], int n, int k)
{
    // increment i in multiples of k where value
    // of k is doubled with each iteration
    for (int i = 0; i < n; i += k/2)
    {
        int left = i;

        // to handle case when number of elements in
        // last group is less than k
        int right = min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr[left++], arr[right--]);

        // double value of k with each iteration
        k = k*2;
    }
}

// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
    int k = 1;

    int n = sizeof(arr) / sizeof(arr[0]);

    reverse(arr, n, k);

    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Java

// Java program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.
import java.util.*;

class GFG
{

// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
static void reverse(int arr[], int n, int k)
{
    // increment i in multiples of k where value
    // of k is doubled with each iteration
    for (int i = 0; i < n; i += k / 2)
    {
        int left = i;

        // to handle case when number of elements in
        // last group is less than k
        int right = Math.min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr, left++, right--);

        // double value of k with each iteration
        k = k * 2;
    }
}

static int[] swap(int[] arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9,
                10, 11, 12, 13, 14, 15, 16};
    int k = 1;

    int n = arr.length;

    reverse(arr, n, k);

    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to reverse every
# sub-array formed by consecutive
# k elements where k doubles itself
# with every sub-array

# Function to reverse every sub-array
# formed by consecutive k elements
# where k doubles itself with every
# sub-array
def reverse(arr, n, k):

    i = 0

    # Increment i in multiples of k where
    # value of k is doubled with each
    # iteration
    while (i < n):
        left = i

        # To handle case when number of elements
        # in last group is less than k
        right = min(i + k - 1, n - 1)

        # Reverse the sub-array [left, right]
        while (left < right):
            arr[left], arr[right] = arr[right], arr[left]
            left += 1
            right -= 1

        # Double value of k with each iteration
        k = k * 2
        i += int(k / 2)

# Driver code
arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9,
        10, 11, 12, 13, 14, 15, 16]
k = 1

n = len(arr)

reverse(arr, n, k)

print(*arr, sep = ' ')

# This code is contributed by avanitrachhadiya2155

C#

// C# program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.
using System;

class GFG
{

// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
static void reverse(int []arr, int n, int k)
{
    // increment i in multiples of k where value
    // of k is doubled with each iteration
    for (int i = 0; i < n; i += k / 2)
    {
        int left = i;

        // to handle case when number of elements in
        // last group is less than k
        int right = Math.Min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr, left++, right--);

        // double value of k with each iteration
        k = k * 2;
    }
}

static int[] swap(int[] arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = {1, 2, 3, 4, 5, 6, 7, 8, 9,
                10, 11, 12, 13, 14, 15, 16};
    int k = 1;

    int n = arr.Length;

    reverse(arr, n, k);

    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.

// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
function reverse(arr,n,k)
{
    // increment i in multiples of k where value
    // of k is doubled with each iteration
    for (let i = 0; i < n; i += k / 2)
    {
        let left = i;

        // to handle case when number of elements in
        // last group is less than k
        let right = Math.min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]
        while (left < right)
            swap(arr, left++, right--);

        // double value of k with each iteration
        k = k * 2;
    }
}

function  swap(arr,i,j)
{
    let temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
let arr=[1, 2, 3, 4, 5, 6, 7, 8, 9,
                10, 11, 12, 13, 14, 15, 16];
let k = 1;
let  n = arr.length;
reverse(arr, n, k);

document.write(arr.join(" "));


// This code is contributed by unknown2108
</script>

Output

1 3 2 7 6 5 4 15 14 13 12 11 10 9 8 16 

Time complexity of all solutions discussed above is O(n). 
Auxiliary space used by the program is O(1).

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.