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Reverse an array in groups of given size

  • Difficulty Level : Easy
  • Last Updated : 25 Jun, 2021

Given an array, reverse every 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, 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] 

Approach: 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 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.

Below image is a dry run of the above approach: 

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>


Output: 

3 2 1 6 5 4 8 7

Time complexity of above solution 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.
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