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++
#include <iostream>
using namespace std;
void reverse( int arr[], int n, int k)
{
for ( int i = 0; i < n; i += k)
{
int left = i;
int right = min(i + k - 1, n - 1);
while (left < right)
swap(arr[left++], arr[right--]);
}
}
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;
}
|
C
#include <stdio.h>
void reverse( int arr[], int n, int k)
{
for ( int i = 0; i < n; i += k)
{
int left = i;
int right;
if (i+k-1<n-1)
right = i+k-1;
else
right = n-1;
while (left < right)
{
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;
}
}
}
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;
}
|
Java
class GFG {
static void reverse( int arr[], int n, int k)
{
for ( int i = 0 ; i < n; i += k)
{
int left = i;
int right = Math.min(i + k - 1 , n - 1 );
int temp;
while (left < right)
{
temp=arr[left];
arr[left]=arr[right];
arr[right]=temp;
left+= 1 ;
right-= 1 ;
}
}
}
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] + " " );
}
}
|
Python3
def reverse(arr, n, k):
i = 0
while (i<n):
left = i
right = min (i + k - 1 , n - 1 )
while (left < right):
arr[left], arr[right] = arr[right], arr[left]
left + = 1 ;
right - = 1
i + = k
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 = " " )
|
C#
using System;
class GFG
{
public static void reverse( int [] arr,
int n, int k)
{
for ( int i = 0; i < n; i += k)
{
int left = i;
int right = Math.Min(i + k - 1, n - 1);
int temp;
while (left < right)
{
temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left += 1;
right -= 1;
}
}
}
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] + " " );
}
}
}
|
Javascript
<script>
function reverse(arr, n, k)
{
for (let i = 0; i < n; i += k)
{
let left = i;
let right = Math.min(i + k - 1, n - 1);
let temp;
while (left < right)
{
temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left += 1;
right -= 1;
}
}
return arr;
}
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] + " " );
</script>
|
PHP
<?php
function reverse( $arr , $n , $k )
{
for ( $i = 0; $i < $n ; $i += $k )
{
$left = $i ;
$right = min( $i + $k - 1, $n - 1);
$temp ;
while ( $left < $right )
{
$temp = $arr [ $left ];
$arr [ $left ] = $arr [ $right ];
$arr [ $right ] = $temp ;
$left += 1;
$right -= 1;
}
}
return $arr ;
}
$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 ] . " " ;
?>
|
Time complexity: O(N)
Auxiliary space: O(1)
Using STL:
C++
#include <bits/stdc++.h>
using namespace std;
void reverse( int arr[], int n, int k)
{
int j = 0, i = k;
while (i < n) {
reverse(arr + j, arr + i);
i += k;
j += k;
}
reverse(arr + j, arr + n);
}
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
import java.util.Arrays;
public class ArrayReverse {
public static void reverse( int [] arr, int n, int k) {
int j = 0 , i = k;
while (i < n) {
reverseArray(arr, j, i - 1 );
i += k;
j += k;
}
reverseArray(arr, j, n - 1 );
}
private static void reverseArray( int [] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
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] + " " );
}
}
|
Python3
def reverse(arr, n, k):
j = 0
i = k
while i < n:
reverse_array(arr, j, i - 1 )
i + = k
j + = k
reverse_array(arr, j, n - 1 )
def reverse_array(arr, start, end):
while start < end:
arr[start], arr[end] = arr[end], arr[start]
start + = 1
end - = 1
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ]
k = 3
n = len (arr)
reverse(arr, n, k)
print ( * arr)
|
C#
using System;
public class ArrayReverse
{
public static void Reverse( int [] arr, int n, int k)
{
int j = 0, i = k;
while (i < n)
{
ReverseArray(arr, j, i - 1);
i += k;
j += k;
}
ReverseArray(arr, j, n - 1);
}
private static void ReverseArray( int [] arr, int start, int end)
{
while (start < end)
{
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
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);
foreach ( int num in arr)
Console.Write(num + " " );
}
}
|
Javascript
function reverse(arr, n, k) {
let j = 0;
let i = k;
while (i < n) {
reverseArray(arr, j, i - 1);
i += k;
j += k;
}
reverseArray(arr, j, n - 1);
}
function reverseArray(arr, start, end) {
while (start < end) {
[arr[start], arr[end]] = [arr[end], arr[start]];
start++;
end--;
}
}
const arr = [1, 2, 3, 4, 5, 6, 7, 8];
const k = 3;
const n = arr.length;
reverse(arr, n, k);
console.log(arr.join( ' ' ));
|
Time complexity: O(N)
Auxiliary space: O(1)
Last Updated :
31 Aug, 2023
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