Reverse an array in groups of given size | Set 2 (Variations of Set 1 )
Last Updated :
23 Jul, 2025
Given an array, reverse every sub-array that satisfies the given constraints.
We have discussed a solution where we reverse every sub-array formed by consecutive k elements in Set 1. In this set, we will discuss various interesting variations of this problem.
Variation 1 (Reverse Alternate Groups): Reverse every alternate 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, 4, 5, 6, 9, 8, 7]
Input:
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 2
Output:
[2, 1, 3, 4, 6, 5, 7, 8]
Algorithm:
- Define a function reverse() that takes an integer array arr, its size n, and the size of the sub-arrays k as input
- Traverse the array in multiples of 2k starting from the first element, i.e., for i = 0, 2k, 4k, and so on.
- For each such i, set the left pointer to i and the right pointer to min(i + k - 1, n - 1) to handle the case when 2k is not a multiple of n.
- Reverse the sub-array [left, right] using a while loop and the swap() function.
- Repeat steps 3-4 for every alternate sub-array formed by consecutive k elements.
- In the main function, declare an integer array arr, initialize it with some values, and define the size of the sub-arrays k.
- Determine the size of the array n using the sizeof() operator.
- Call the reverse() function passing the integer array arr, its size n, and the size of the sub-arrays k as input.
- Print the modified array arr.
Below is the implementation :
C++
// C++ program to reverse every alternate sub-array
// formed by consecutive k elements
#include <iostream>
using namespace std;
// Function to reverse every alternate sub-array
// formed by consecutive k elements
void reverse(int arr[], int n, int k)
{
// increment i in multiples of 2*k
for (int i = 0; i < n; i += 2*k)
{
int left = i;
// to handle case when 2*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, 9, 10, 11, 12, 13, 14};
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 program to reverse
// every alternate sub-array
// formed by consecutive k elements
class GFG
{
// Function to reverse every
// alternate sub-array formed
// by consecutive k elements
static void reverse(int arr[], int n, int k)
{
// increment i in multiples of 2*k
for (int i = 0; i < n; i += 2 * k)
{
int left = i;
// to handle case when 2*k is not multiple of n
int right = Math.min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right) {
swap(arr, left++, right--);
}
}
}
static int[] swap(int[] array, int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
return array;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14};
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 has been contributed by 29AjayKumar
# Python3 program to reverse every alternate sub-array
# formed by consecutive k elements
# Function to reverse every alternate sub-array
# formed by consecutive k elements
def reverse(arr, n, k):
# increment i in multiples of 2*k
for i in range(0,n,2*k):
left = i
# to handle case when 2*k is not multiple of n
right = min(i + k - 1, n - 1)
# 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
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
k = 3
n = len(arr)
reverse(arr, n, k)
for i in range(0,n,1):
print(arr[i],end =" ")
# This code is contributed by
# Surendra_Gangwar
// C# program to reverse every alternate
// sub-array formed by consecutive k elements
using System;
class GFG
{
// Function to reverse every
// alternate sub-array formed
// by consecutive k elements
static void reverse(int []arr,
int n, int k)
{
// increment i in multiples of 2*k
for (int i = 0; i < n; i += 2 * k)
{
int left = i;
// to handle case when 2*k is
// not multiple of n
int right = Math.Min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
{
swap(arr, left++, right--);
}
}
}
static int[] swap(int[] array, int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
return array;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14};
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 PrinciRaj1992
<script>
// Javascript program to reverse
// every alternate sub-array
// formed by consecutive k elements
// Function to reverse every
// alternate sub-array formed
// by consecutive k elements
function reverse(arr, n, k)
{
// Increment i in multiples of 2*k
for(let i = 0; i < n; i += 2 * k)
{
let left = i;
// To handle case when 2*k is
// not multiple of n
let right = Math.min(i + k - 1,
n - 1);
// reverse the sub-array [left, right]
while (left < right)
{
swap(arr, left++, right--);
}
}
}
function swap(array, i, j)
{
let temp = array[i];
array[i] = array[j];
array[j] = temp;
return array;
}
// Driver code
let arr = [ 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14 ];
let k = 3;
let n = arr.length;
reverse(arr, n, k);
for(let i = 0; i < n; i++)
{
document.write(arr[i] + " ");
}
// This code is contributed by rag2127
</script>
Output3 2 1 4 5 6 9 8 7 10 11 12 14 13
Time Complexity: O(N)
Auxiliary Space: O(1)
Variation 2 (Reverse at given distance): Reverse every sub-array formed by consecutive k elements at given distance apart.
Examples:
Input:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
k = 3
m = 2
Output:
[3, 2, 1, 4, 5, 8, 7, 6, 9, 10]
Input:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
k = 3
m = 1
Output:
[3, 2, 1, 4, 7, 6, 5, 8, 10, 9]
Input:
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 2
m = 0
Output:
[2, 1, 4, 3, 6, 5, 8, 7]
Below is its implementation:
C++
// C++ program to reverse every sub-array formed by
// consecutive k elements at given distance apart
#include <iostream>
using namespace std;
// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
void reverse(int arr[], int n, int k, int m)
{
// increment i in multiples of k + m
for (int i = 0; i < n; i += k + m)
{
int left = i;
// to handle case when k + m 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, 9, 10, 11, 12, 13, 14};
int k = 3;
int m = 2;
int n = sizeof(arr) / sizeof(arr[0]);
reverse(arr, n, k, m);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
// java program to reverse every sub-array formed by
// consecutive k elements at given distance apart
class GFG
{
// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
static void reverse(int[] arr, int n, int k, int m)
{
// increment i in multiples of k + m
for (int i = 0; i < n; i += k + m)
{
int left = i;
// to handle case when k + m is not multiple of n
int right = Math.min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr,left++, right--);
}
}
static int[] swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14};
int k = 3;
int m = 2;
int n = arr.length;
reverse(arr, n, k, m );
for (int i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
}
}
// This code has been contributed by Rajput-Ji
# Python3 program to reverse every
# sub-array formed by consecutive
# k elements at given distance apart
# Function to reverse every
# sub-array formed by consecutive
# k elements at m distance apart
def reverse(arr, n, k, m):
# increment i in multiples of k + m
for i in range(0, n, k + m):
left = i;
# to handle case when k + m
# is not multiple of n
right = min(i + k - 1, n - 1);
# reverse the sub-array [left, right]
while (left < right):
arr = swap(arr,left, right);
left += 1;
right -= 1;
return arr;
def swap(arr, i, j):
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14];
k = 3;
m = 2;
n = len(arr);
arr = reverse(arr, n, k, m );
for i in range(0, n):
print(arr[i], end = " ");
# This code is contributed by Rajput-Ji
// C# program to reverse every sub-array
// formed by consecutive k elements at
// given distance apart
using System;
class GFG
{
// Function to reverse every sub-array
// formed by consecutive k elements
// at m distance apart
static void reverse(int[] arr, int n,
int k, int m)
{
// increment i in multiples of k + m
for (int i = 0; i < n; i += k + m)
{
int left = i;
// to handle case when k + m is
// not multiple of n
int right = Math.Min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr, left++, right--);
}
}
static int[] swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14};
int k = 3;
int m = 2;
int n = arr.Length;
reverse(arr, n, k, m );
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
}
// This code is contributed by PrinciRaj1992
<script>
// javascript program to reverse every sub-array formed by
// consecutive k elements at given distance apart
// Function to reverse every sub-array formed by
// consecutive k elements at m distance apart
function reverse(arr,n,k,m)
{
// increment i in multiples of k + m
for (let i = 0; i < n; i += k + m)
{
let left = i;
// to handle case when k + m is not multiple of n
let right = Math.min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr,left++, right--);
}
}
function swap(arr,i,j)
{
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
let arr=[1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14];
let k = 3;
let m = 2;
let n = arr.length;
reverse(arr, n, k, m );
for (let i = 0; i < n; i++)
{
document.write(arr[i] + " ");
}
// This code is contributed by ab2127
</script>
Output3 2 1 4 5 8 7 6 9 10 13 12 11 14
Time Complexity: O(N)
Auxiliary Space: O(1)
Variation 3 (Reverse by doubling the group size):
Reverse every sub-array formed by consecutive k elements where k doubles itself with every sub-array.
Examples:
Input:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
k = 1
Output:
[1], [3, 2], [7, 6, 5, 4], [15, 14, 13, 12, 11, 10, 9, 8]
Below is its implementation:
C++
// C++ program to reverse every sub-array formed by
// consecutive k elements where k doubles itself with
// every sub-array.
#include <iostream>
using namespace std;
// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
void reverse(int arr[], int n, int k)
{
// increment i in multiples of k where value
// of k is doubled with each iteration
for (int i = 0; i < n; i += k/2)
{
int left = i;
// to handle case when number of elements in
// last group is less than k
int right = min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr[left++], arr[right--]);
// double value of k with each iteration
k = k*2;
}
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
int k = 1;
int n = sizeof(arr) / sizeof(arr[0]);
reverse(arr, n, k);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
// Java program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.
import java.util.*;
class GFG
{
// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
static void reverse(int arr[], int n, int k)
{
// increment i in multiples of k where value
// of k is doubled with each iteration
for (int i = 0; i < n; i += k / 2)
{
int left = i;
// to handle case when number of elements in
// last group is less than k
int right = Math.min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr, left++, right--);
// double value of k with each iteration
k = k * 2;
}
}
static int[] swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16};
int k = 1;
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 29AjayKumar
# Python3 program to reverse every
# sub-array formed by consecutive
# k elements where k doubles itself
# with every sub-array
# Function to reverse every sub-array
# formed by consecutive k elements
# where k doubles itself with every
# sub-array
def reverse(arr, n, k):
i = 0
# Increment i in multiples of k where
# value of k is doubled with each
# iteration
while (i < n):
left = i
# To handle case when number of elements
# in last group is less than k
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
# Double value of k with each iteration
k = k * 2
i += int(k / 2)
# Driver code
arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16]
k = 1
n = len(arr)
reverse(arr, n, k)
print(*arr, sep = ' ')
# This code is contributed by avanitrachhadiya2155
// C# program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.
using System;
class GFG
{
// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
static void reverse(int []arr, int n, int k)
{
// increment i in multiples of k where value
// of k is doubled with each iteration
for (int i = 0; i < n; i += k / 2)
{
int left = i;
// to handle case when number of elements in
// last group is less than k
int right = Math.Min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr, left++, right--);
// double value of k with each iteration
k = k * 2;
}
}
static int[] swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16};
int k = 1;
int n = arr.Length;
reverse(arr, n, k);
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program to reverse every sub-array
// formed by consecutive k elements where
// k doubles itself with every sub-array.
// Function to reverse every sub-array formed by
// consecutive k elements where k doubles itself
// with every sub-array.
function reverse(arr,n,k)
{
// increment i in multiples of k where value
// of k is doubled with each iteration
for (let i = 0; i < n; i += k / 2)
{
let left = i;
// to handle case when number of elements in
// last group is less than k
let right = Math.min(i + k - 1, n - 1);
// reverse the sub-array [left, right]
while (left < right)
swap(arr, left++, right--);
// double value of k with each iteration
k = k * 2;
}
}
function swap(arr,i,j)
{
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
let arr=[1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16];
let k = 1;
let n = arr.length;
reverse(arr, n, k);
document.write(arr.join(" "));
// This code is contributed by unknown2108
</script>
Output1 3 2 7 6 5 4 15 14 13 12 11 10 9 8 16
Time complexity of all solutions discussed above is O(n).
Auxiliary space used by the program is O(1).
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