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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: 

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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).

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