Skip to content
Related Articles

Related Articles

Reverse an Array in groups of given size

Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 29 Oct, 2022
Improve Article
Save Article

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

Recommended Practice

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)

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.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!