Reverse an array in groups of given size

Software Development
#1

Hello Everyone,

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 is the implementation of the above approach:

// 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;

}

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).