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# Print Array after it is right rotated K times where K can be large or negative

• Last Updated : 21 Sep, 2022

Given an array arr[] of size N and a value K (-10^5<K<10^5), the task is to print the array rotated by K times to the right.

Examples:

Input: arr = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5
Explanation:
Rotating array 1 time right: 9, 1, 3, 5, 7
Rotating array 2 time right: 7, 9, 1, 3, 5

Input: arr = {1, 2, 3, 4, 5}, K = -2
Output: 3 4 5 1 2
Explanation:
Rotating array -1 time right: 2, 3, 4, 5, 1
Rotating array -2 time right: 3, 4, 5, 1, 2

Naive Approach: The brute force approach to solve this problem is to use a temporary array to rotate the array K or -K times.

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach: The given problem can be solved by breaking the problem into the following parts:

1. Round up the value of K in range [0, N), using below steps:
• If K is negative, first change it into positive, find the modulo with N, and then again change it to negative
• If K is positive, just find the modulo with N
2. Handle the case when K is negative. If K is negative, it means we need to rotate the array K times left, or -K times right.
3. Next we can simply rotate the array K times by reversing subarrays. Below steps can be followed to solve the problem:
• Reverse all the array elements from 1 to N -1
• Reverse the array elements from 1 to K â€“ 1
• Reverse the array elements from K to N -1

## C++

```// C++ implementation for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to rotate the array
// to the right, K times
void RightRotate(vector<int>& nums, int K)
{
int n = nums.size();

// Case when K > N or K < -N
K = K < 0 ? ((K * -1) % n) * -1 : K % n;

// Case when K is negative
K = K < 0 ? (n - (K * -1)) : K;

// Reverse all the array elements
reverse(nums.begin(), nums.end());

// Reverse the first k elements
reverse(nums.begin(), nums.begin() + K);

// Reverse the elements from K
// till the end of the array
reverse(nums.begin() + K, nums.end());
}

// Driver code
int main()
{

// Initialize the array
vector<int> Array = { 1, 2, 3, 4, 5 };

// Find the size of the array
int N = Array.size();

// Initialize K
int K = -2;

// Call the function and
RightRotate(Array, K);

// Print the array after rotation
for (int i = 0; i < N; i++) {

cout << Array[i] << " ";
}

cout << endl;
return 0;
}```

## Java

```// Java implementation for the above approach
import java.util.*;

class GFG{

// Initialize the array
static int[] Array = { 1, 2, 3, 4, 5 };

static void reverse( int start, int end) {

// Temporary variable to store character
int temp;
while (start <= end)
{

// Swapping the first and last character
temp = Array[start];
Array[start] = Array[end];
Array[end] = temp;
start++;
end--;
}
}

// Function to rotate the array
// to the right, K times
static void RightRotate( int K)
{
int n = Array.length;

// Case when K > N or K < -N
K = K < 0 ? ((K * -1) % n) * -1 : K % n;

// Case when K is negative
K = K < 0 ? (n - (K * -1)) : K;

// Reverse all the array elements
reverse(0, n-1);

// Reverse the first k elements
reverse(0, n - K);

// Reverse the elements from K
// till the end of the array
reverse( K, n-1);
}

// Driver code
public static void main(String[] args)
{

// Find the size of the array
int N = Array.length;

// Initialize K
int K = -2;

// Call the function and
RightRotate(K);

// Print the array after rotation
for (int i = 0; i < N; i++) {

System.out.print(Array[i]+ " ");
}

System.out.println();
}
}

// This code is contributed by Rajput-Ji
```

## Python3

```# Python code for the above approach

# Function to rotate the array
# to the right, K times
def RightRotate(nums, K) :
n = len(nums)

# Case when K > N or K < -N
K = ((K * -1) % n) * -1 if K < 0 else K % n;

# Case when K is negative
K = (n - (K * -1)) if K < 0 else K;

# Reverse all the array elements
nums.reverse();

# Reverse the first k elements
p1 = nums[0:K]
p1.reverse();

# Reverse the elements from K
# till the end of the array
p2 = nums[K:]
p2.reverse();
arr = p1 + p2

return arr;

# Driver code

# Initialize the array
Array = [1, 2, 3, 4, 5];

# Find the size of the array
N = len(Array)

# Initialize K
K = -2;

# Call the function and
Array = RightRotate(Array, K);

# Print the array after rotation
for i in Array:
print(i, end=" ")

# This code is contributed by Saurabh jaiswal```

## C#

```// C# implementation for the above approach
using System;
public class GFG {

// Initialize the array
static int[] Array = { 1, 2, 3, 4, 5 };
static void reverse(int start, int end)
{

// Temporary variable to store character
int temp;
while (start <= end) {

// Swapping the first and last character
temp = Array[start];
Array[start] = Array[end];
Array[end] = temp;
start++;
end--;
}
}

// Function to rotate the array
// to the right, K times
static void RightRotate(int K) {
int n = Array.Length;

// Case when K > N or K < -N
K = K < 0 ? ((K * -1) % n) * -1 : K % n;

// Case when K is negative
K = K < 0 ? (n - (K * -1)) : K;

// Reverse all the array elements
reverse(0, n - 1);

// Reverse the first k elements
reverse(0, n - K);

// Reverse the elements from K
// till the end of the array
reverse(K, n - 1);
}

// Driver code
public static void Main(String[] args) {

// Find the size of the array
int N = Array.Length;

// Initialize K
int K = -2;

// Call the function and
RightRotate(K);

// Print the array after rotation
for (int i = 0; i < N; i++) {

Console.Write(Array[i] + " ");
}

Console.WriteLine();
}
}

// This code is contributed by Rajput-Ji
```

## Javascript

``` <script>
// JavaScript code for the above approach

// Function to rotate the array
// to the right, K times
function RightRotate(nums, K)
{
let n = nums.length;

// Case when K > N or K < -N
K = K < 0 ? ((K * -1) % n) * -1 : K % n;

// Case when K is negative
K = K < 0 ? (n - (K * -1)) : K;

// Reverse all the array elements
nums = nums.reverse();

// Reverse the first k elements
let p1 = nums.slice(0, K)
p1 = p1.reverse();

// Reverse the elements from K
// till the end of the array
let p2 = nums.slice(K)
p2 = p2.reverse();

let arr = p1.concat(p2);

return arr;
}

// Driver code

// Initialize the array
let Array = [1, 2, 3, 4, 5];

// Find the size of the array
let N = Array.length;

// Initialize K
let K = -2;

// Call the function and
Array = RightRotate(Array, K);

// Print the array after rotation
for (let i = 0; i < N; i++) {

document.write(Array[i] + " ");
}
document.write('<br>')

// This code is contributed by Potta Lokesh
</script>```

Output

`3 4 5 1 2 `

Time Complexity: O(N)
Auxiliary Space: O(1)

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