Reverse an Array in groups of given size

Given an array arr[] and an integer K, the task is to reverse every subarray formed by consecutive K elements.

Examples:

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

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

Input: arr[] = [1, 2, 3, 4, 5, 6], K = 1 
Output: 1, 2, 3, 4, 5, 6

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

Naive Approach: The problem can be solved based on the following idea:

Consider every sub-array of size k starting from the beginning of the array and reverse it. We need to handle some special cases. 

• If k is not a multiple of n where n is the size of the array, for the last group we will have less than k elements left, we need to reverse all remaining elements. 
• If k = 1, the array should remain unchanged. If k >= n, we reverse all elements present in the array.

Follow the below illustration for a better understanding.

Illustration:

Given array arr[] = {1, 2, 3, 4, 5, 6, 7, 8} and k = 3

1st iteration:

• The selected group is {1, 2, 3, 4, 5, 6, 7, 8}
• After swapping the elements we have {3, 2, 1, 4, 5, 6, 7, 8}

2nd iteration: 

• The selected group is {3, 2, 1, 4, 5, 6, 7, 8}
• After swapping the elements {3, 2, 1, 6, 5, 4, 7, 8}

3rd iteration: 

• As k is less than the count of remaining elements
• We will reverse the entire remaining subarray {3, 2, 1, 6, 5, 4, 8, 7}

Follow the steps mentioned below to implement the idea:

• Iterate over the array, and on each iteration:
  • We will set the left pointer as the current index and the right pointer at a distance of group size(K) 
  • We will swap elements in the left and right indices, and increase left by one and decrease right by one
  • Do the above step until left < right
  • After the swap operation is done we will increment the value of the iterator by k ( group size )

Below is the implementation of the above approach:

C++

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

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

        // to handle case when 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};
    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 sub-array formed by
// consecutive k elements
class GFG {

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

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

            // reverse the sub-array [left, right]
            while (left < right)
            {
                temp=arr[left];
                arr[left]=arr[right];
                arr[right]=temp;
                left+=1;
                right-=1;
            }
        }
    }

    // Driver method
    public static void main(String[] args)
    {

        int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
        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 is contributed by Anant Agarwal.

Python3

# Python 3 program to reverse every
# sub-array formed by consecutive k
# elements

# Function to reverse every sub-array
# formed by consecutive k elements
def reverse(arr, n, k):
    i = 0

    while(i<n):

        left = i

        # To handle case when k is not
        # multiple of n
        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
        i+= k

# Driver code
arr = [1, 2, 3, 4, 5, 6,
                   7, 8]

k = 3
n = len(arr)
reverse(arr, n, k)

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

# This code is contributed by Smitha Dinesh Semwal

C#

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

class GFG
{

// Function to reverse every sub-array
// formed by consecutive k elements
public static void reverse(int[] arr,
                           int n, int k)
{
    for (int i = 0; i < n; i += k)
    {
        int left = i;

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

        // 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
public static void Main(string[] args)
{
    int[] arr = new int[] {1, 2, 3, 4,
                           5, 6, 7, 8};
    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 Shrikant13

PHP

<?php
// PHP program to reverse every sub-array
// formed by consecutive k elements

// Function to reverse every sub-array
// formed by consecutive k elements
function reverse($arr, $n, $k)
{
    for ($i = 0; $i < $n; $i += $k)
    {
        $left = $i;

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

        // reverse the sub-array [left, right]
        while ($left < $right)
        {
            $temp = $arr[$left];
            $arr[$left] = $arr[$right];
            $arr[$right] = $temp;
            $left += 1;
            $right -= 1;
        }
    }
    return $arr;
}

// Driver Code
$arr = array(1, 2, 3, 4, 5, 6, 7, 8);
$k = 3;

$n = sizeof($arr);

$arr1 = reverse($arr, $n, $k);

for ($i = 0; $i < $n; $i++)
    echo $arr1[$i] . " ";

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

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

// Function to reverse every sub-array
// formed by consecutive k elements
function reverse(arr, n, k)
{
    for(let i = 0; i < n; i += k)
    {
        let left = i;

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

        // Reverse the sub-array [left, right]
        while (left < right)
        {
            temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
            left += 1;
            right -= 1;
        }
    }
    return arr;
}

// Driver Code
let arr = new Array(1, 2, 3, 4, 5, 6, 7, 8);
let k = 3;
let n = arr.length;
let arr1 = reverse(arr, n, k);

for(let i = 0; i < n; i++)
    document.write(arr1[i] + " ");

// This code is contributed by saurabh jaiswal

</script>

C

// C program to reverse every sub-array formed by
// consecutive k elements
#include <stdio.h>
// Function to reverse every sub-array formed by
// consecutive k elements
void reverse(int arr[], int n, int k)
{
    for (int i = 0; i < n; i += k)
    {
        int left = i;
        int right;
        // to handle case when k is not multiple of n
        if(i+k-1<n-1)
        right = i+k-1;
        else
        right = n-1;

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

    }
}

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

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

    reverse(arr, n, k);

    for (int i = 0; i < n; i++)
        printf("%d ",arr[i]);

    return 0;
}
//  This code is contributed by Arpit Jain

Output

3 2 1 6 5 4 8 7 

Time complexity: O(N)
Auxiliary space: O(1)

Using STL:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to reverse subarrays of size k in an array
// Nikunj Sonigara
void reverse(int arr[], int n, int k)
{
    int j = 0, i = k;

    // Reverse subarrays of size k until the end of the array is reached
    while (i < n) {
        // Reverse the subarray from arr[j] to arr[i-1]
        reverse(arr + j, arr + i);

        // Move to the next subarray of size k
        i += k;
        j += k;
    }

    // Reverse the remaining subarray of size less than k
    reverse(arr + j, arr + n);
}

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

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

    // Reverse subarrays of size k in the array
    reverse(arr, n, k);

    // Print the reversed array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";

    return 0;
}

Output

3 2 1 6 5 4 8 7 

Time complexity: O(N)
Auxiliary space: O(1)

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